(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Begin(5(5(5(5(5(4(4(4(4(4(4(x)))))))))))) → Wait(Right4(x))
Begin(5(5(5(5(4(4(4(4(4(4(x))))))))))) → Wait(Right5(x))
Begin(5(5(5(4(4(4(4(4(4(x)))))))))) → Wait(Right6(x))
Begin(5(5(4(4(4(4(4(4(x))))))))) → Wait(Right7(x))
Begin(5(4(4(4(4(4(4(x)))))))) → Wait(Right8(x))
Begin(4(4(4(4(4(4(x))))))) → Wait(Right9(x))
Begin(4(4(4(4(4(x)))))) → Wait(Right10(x))
Begin(4(4(4(4(x))))) → Wait(Right11(x))
Begin(4(4(4(x)))) → Wait(Right12(x))
Begin(4(4(x))) → Wait(Right13(x))
Begin(4(x)) → Wait(Right14(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right4(5(End(x))) → Left(2(End(x)))
Right5(5(5(End(x)))) → Left(2(End(x)))
Right6(5(5(5(End(x))))) → Left(2(End(x)))
Right7(5(5(5(5(End(x)))))) → Left(2(End(x)))
Right8(5(5(5(5(5(End(x))))))) → Left(2(End(x)))
Right9(5(5(5(5(5(5(End(x)))))))) → Left(2(End(x)))
Right10(5(5(5(5(5(5(4(End(x))))))))) → Left(2(End(x)))
Right11(5(5(5(5(5(5(4(4(End(x)))))))))) → Left(2(End(x)))
Right12(5(5(5(5(5(5(4(4(4(End(x))))))))))) → Left(2(End(x)))
Right13(5(5(5(5(5(5(4(4(4(4(End(x)))))))))))) → Left(2(End(x)))
Right14(5(5(5(5(5(5(4(4(4(4(4(End(x))))))))))))) → Left(2(End(x)))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
0(x) → 1(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
5(5(5(5(5(5(4(4(4(4(4(4(x)))))))))))) → 2(x)
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0(x1)) = 1 + x1
POL(1(x1)) = x1
POL(2(x1)) = 3 + x1
POL(4(x1)) = x1
POL(5(x1)) = 1 + x1
POL(A0(x1)) = 1 + x1
POL(A1(x1)) = x1
POL(A2(x1)) = 3 + x1
POL(A4(x1)) = x1
POL(A5(x1)) = 1 + x1
POL(Begin(x1)) = 1 + x1
POL(End(x1)) = x1
POL(Left(x1)) = 1 + x1
POL(Right1(x1)) = 3 + x1
POL(Right10(x1)) = x1
POL(Right11(x1)) = x1
POL(Right12(x1)) = x1
POL(Right13(x1)) = x1
POL(Right14(x1)) = x1
POL(Right2(x1)) = 2 + x1
POL(Right3(x1)) = 2 + x1
POL(Right4(x1)) = 4 + x1
POL(Right5(x1)) = 4 + x1
POL(Right6(x1)) = 2 + x1
POL(Right7(x1)) = 1 + x1
POL(Right8(x1)) = x1
POL(Right9(x1)) = x1
POL(Wait(x1)) = x1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
Begin(5(5(5(5(5(4(4(4(4(4(4(x)))))))))))) → Wait(Right4(x))
Begin(5(5(5(5(4(4(4(4(4(4(x))))))))))) → Wait(Right5(x))
Begin(5(5(5(4(4(4(4(4(4(x)))))))))) → Wait(Right6(x))
Begin(5(5(4(4(4(4(4(4(x))))))))) → Wait(Right7(x))
Begin(5(4(4(4(4(4(4(x)))))))) → Wait(Right8(x))
Begin(4(4(4(4(4(4(x))))))) → Wait(Right9(x))
Begin(4(4(4(4(4(x)))))) → Wait(Right10(x))
Begin(4(4(4(4(x))))) → Wait(Right11(x))
Begin(4(4(4(x)))) → Wait(Right12(x))
Begin(4(4(x))) → Wait(Right13(x))
Begin(4(x)) → Wait(Right14(x))
Right4(5(End(x))) → Left(2(End(x)))
Right5(5(5(End(x)))) → Left(2(End(x)))
Right6(5(5(5(End(x))))) → Left(2(End(x)))
Right7(5(5(5(5(End(x)))))) → Left(2(End(x)))
Right8(5(5(5(5(5(End(x))))))) → Left(2(End(x)))
Right9(5(5(5(5(5(5(End(x)))))))) → Left(2(End(x)))
Right10(5(5(5(5(5(5(4(End(x))))))))) → Left(2(End(x)))
Right11(5(5(5(5(5(5(4(4(End(x)))))))))) → Left(2(End(x)))
Right12(5(5(5(5(5(5(4(4(4(End(x))))))))))) → Left(2(End(x)))
Right13(5(5(5(5(5(5(4(4(4(4(End(x)))))))))))) → Left(2(End(x)))
Right14(5(5(5(5(5(5(4(4(4(4(4(End(x))))))))))))) → Left(2(End(x)))
0(x) → 1(x)
5(5(5(5(5(5(4(4(4(4(4(4(x)))))))))))) → 2(x)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(5(4(5(x)))) → RIGHT1(x)
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(4(5(x))) → RIGHT2(x)
BEGIN(5(x)) → WAIT(Right3(x))
BEGIN(5(x)) → RIGHT3(x)
RIGHT1(4(End(x))) → 41(4(5(5(End(x)))))
RIGHT1(4(End(x))) → 41(5(5(End(x))))
RIGHT2(4(5(End(x)))) → 41(4(5(5(End(x)))))
RIGHT2(4(5(End(x)))) → 41(5(5(End(x))))
RIGHT3(4(5(4(End(x))))) → 41(4(5(5(End(x)))))
RIGHT3(4(5(4(End(x))))) → 41(5(5(End(x))))
RIGHT1(0(x)) → A01(Right1(x))
RIGHT1(0(x)) → RIGHT1(x)
RIGHT2(0(x)) → A01(Right2(x))
RIGHT2(0(x)) → RIGHT2(x)
RIGHT3(0(x)) → A01(Right3(x))
RIGHT3(0(x)) → RIGHT3(x)
RIGHT4(0(x)) → A01(Right4(x))
RIGHT4(0(x)) → RIGHT4(x)
RIGHT5(0(x)) → A01(Right5(x))
RIGHT5(0(x)) → RIGHT5(x)
RIGHT6(0(x)) → A01(Right6(x))
RIGHT6(0(x)) → RIGHT6(x)
RIGHT7(0(x)) → A01(Right7(x))
RIGHT7(0(x)) → RIGHT7(x)
RIGHT8(0(x)) → A01(Right8(x))
RIGHT8(0(x)) → RIGHT8(x)
RIGHT9(0(x)) → A01(Right9(x))
RIGHT9(0(x)) → RIGHT9(x)
RIGHT10(0(x)) → A01(Right10(x))
RIGHT10(0(x)) → RIGHT10(x)
RIGHT11(0(x)) → A01(Right11(x))
RIGHT11(0(x)) → RIGHT11(x)
RIGHT12(0(x)) → A01(Right12(x))
RIGHT12(0(x)) → RIGHT12(x)
RIGHT13(0(x)) → A01(Right13(x))
RIGHT13(0(x)) → RIGHT13(x)
RIGHT14(0(x)) → A01(Right14(x))
RIGHT14(0(x)) → RIGHT14(x)
RIGHT1(1(x)) → A11(Right1(x))
RIGHT1(1(x)) → RIGHT1(x)
RIGHT2(1(x)) → A11(Right2(x))
RIGHT2(1(x)) → RIGHT2(x)
RIGHT3(1(x)) → A11(Right3(x))
RIGHT3(1(x)) → RIGHT3(x)
RIGHT4(1(x)) → A11(Right4(x))
RIGHT4(1(x)) → RIGHT4(x)
RIGHT5(1(x)) → A11(Right5(x))
RIGHT5(1(x)) → RIGHT5(x)
RIGHT6(1(x)) → A11(Right6(x))
RIGHT6(1(x)) → RIGHT6(x)
RIGHT7(1(x)) → A11(Right7(x))
RIGHT7(1(x)) → RIGHT7(x)
RIGHT8(1(x)) → A11(Right8(x))
RIGHT8(1(x)) → RIGHT8(x)
RIGHT9(1(x)) → A11(Right9(x))
RIGHT9(1(x)) → RIGHT9(x)
RIGHT10(1(x)) → A11(Right10(x))
RIGHT10(1(x)) → RIGHT10(x)
RIGHT11(1(x)) → A11(Right11(x))
RIGHT11(1(x)) → RIGHT11(x)
RIGHT12(1(x)) → A11(Right12(x))
RIGHT12(1(x)) → RIGHT12(x)
RIGHT13(1(x)) → A11(Right13(x))
RIGHT13(1(x)) → RIGHT13(x)
RIGHT14(1(x)) → A11(Right14(x))
RIGHT14(1(x)) → RIGHT14(x)
RIGHT1(4(x)) → A41(Right1(x))
RIGHT1(4(x)) → RIGHT1(x)
RIGHT2(4(x)) → A41(Right2(x))
RIGHT2(4(x)) → RIGHT2(x)
RIGHT3(4(x)) → A41(Right3(x))
RIGHT3(4(x)) → RIGHT3(x)
RIGHT4(4(x)) → A41(Right4(x))
RIGHT4(4(x)) → RIGHT4(x)
RIGHT5(4(x)) → A41(Right5(x))
RIGHT5(4(x)) → RIGHT5(x)
RIGHT6(4(x)) → A41(Right6(x))
RIGHT6(4(x)) → RIGHT6(x)
RIGHT7(4(x)) → A41(Right7(x))
RIGHT7(4(x)) → RIGHT7(x)
RIGHT8(4(x)) → A41(Right8(x))
RIGHT8(4(x)) → RIGHT8(x)
RIGHT9(4(x)) → A41(Right9(x))
RIGHT9(4(x)) → RIGHT9(x)
RIGHT10(4(x)) → A41(Right10(x))
RIGHT10(4(x)) → RIGHT10(x)
RIGHT11(4(x)) → A41(Right11(x))
RIGHT11(4(x)) → RIGHT11(x)
RIGHT12(4(x)) → A41(Right12(x))
RIGHT12(4(x)) → RIGHT12(x)
RIGHT13(4(x)) → A41(Right13(x))
RIGHT13(4(x)) → RIGHT13(x)
RIGHT14(4(x)) → A41(Right14(x))
RIGHT14(4(x)) → RIGHT14(x)
RIGHT1(5(x)) → A51(Right1(x))
RIGHT1(5(x)) → RIGHT1(x)
RIGHT2(5(x)) → A51(Right2(x))
RIGHT2(5(x)) → RIGHT2(x)
RIGHT3(5(x)) → A51(Right3(x))
RIGHT3(5(x)) → RIGHT3(x)
RIGHT4(5(x)) → A51(Right4(x))
RIGHT4(5(x)) → RIGHT4(x)
RIGHT5(5(x)) → A51(Right5(x))
RIGHT5(5(x)) → RIGHT5(x)
RIGHT6(5(x)) → A51(Right6(x))
RIGHT6(5(x)) → RIGHT6(x)
RIGHT7(5(x)) → A51(Right7(x))
RIGHT7(5(x)) → RIGHT7(x)
RIGHT8(5(x)) → A51(Right8(x))
RIGHT8(5(x)) → RIGHT8(x)
RIGHT9(5(x)) → A51(Right9(x))
RIGHT9(5(x)) → RIGHT9(x)
RIGHT10(5(x)) → A51(Right10(x))
RIGHT10(5(x)) → RIGHT10(x)
RIGHT11(5(x)) → A51(Right11(x))
RIGHT11(5(x)) → RIGHT11(x)
RIGHT12(5(x)) → A51(Right12(x))
RIGHT12(5(x)) → RIGHT12(x)
RIGHT13(5(x)) → A51(Right13(x))
RIGHT13(5(x)) → RIGHT13(x)
RIGHT14(5(x)) → A51(Right14(x))
RIGHT14(5(x)) → RIGHT14(x)
RIGHT1(2(x)) → A21(Right1(x))
RIGHT1(2(x)) → RIGHT1(x)
RIGHT2(2(x)) → A21(Right2(x))
RIGHT2(2(x)) → RIGHT2(x)
RIGHT3(2(x)) → A21(Right3(x))
RIGHT3(2(x)) → RIGHT3(x)
RIGHT4(2(x)) → A21(Right4(x))
RIGHT4(2(x)) → RIGHT4(x)
RIGHT5(2(x)) → A21(Right5(x))
RIGHT5(2(x)) → RIGHT5(x)
RIGHT6(2(x)) → A21(Right6(x))
RIGHT6(2(x)) → RIGHT6(x)
RIGHT7(2(x)) → A21(Right7(x))
RIGHT7(2(x)) → RIGHT7(x)
RIGHT8(2(x)) → A21(Right8(x))
RIGHT8(2(x)) → RIGHT8(x)
RIGHT9(2(x)) → A21(Right9(x))
RIGHT9(2(x)) → RIGHT9(x)
RIGHT10(2(x)) → A21(Right10(x))
RIGHT10(2(x)) → RIGHT10(x)
RIGHT11(2(x)) → A21(Right11(x))
RIGHT11(2(x)) → RIGHT11(x)
RIGHT12(2(x)) → A21(Right12(x))
RIGHT12(2(x)) → RIGHT12(x)
RIGHT13(2(x)) → A21(Right13(x))
RIGHT13(2(x)) → RIGHT13(x)
RIGHT14(2(x)) → A21(Right14(x))
RIGHT14(2(x)) → RIGHT14(x)
A41(Left(x)) → 41(x)
WAIT(Left(x)) → BEGIN(x)
41(5(4(5(x)))) → 41(4(5(5(x))))
41(5(4(5(x)))) → 41(5(5(x)))
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 15 SCCs with 82 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT14(1(x)) → RIGHT14(x)
RIGHT14(0(x)) → RIGHT14(x)
RIGHT14(4(x)) → RIGHT14(x)
RIGHT14(5(x)) → RIGHT14(x)
RIGHT14(2(x)) → RIGHT14(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT14(1(x)) → RIGHT14(x)
RIGHT14(0(x)) → RIGHT14(x)
RIGHT14(4(x)) → RIGHT14(x)
RIGHT14(5(x)) → RIGHT14(x)
RIGHT14(2(x)) → RIGHT14(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT14(1(x)) → RIGHT14(x)
The graph contains the following edges 1 > 1
- RIGHT14(0(x)) → RIGHT14(x)
The graph contains the following edges 1 > 1
- RIGHT14(4(x)) → RIGHT14(x)
The graph contains the following edges 1 > 1
- RIGHT14(5(x)) → RIGHT14(x)
The graph contains the following edges 1 > 1
- RIGHT14(2(x)) → RIGHT14(x)
The graph contains the following edges 1 > 1
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT13(1(x)) → RIGHT13(x)
RIGHT13(0(x)) → RIGHT13(x)
RIGHT13(4(x)) → RIGHT13(x)
RIGHT13(5(x)) → RIGHT13(x)
RIGHT13(2(x)) → RIGHT13(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT13(1(x)) → RIGHT13(x)
RIGHT13(0(x)) → RIGHT13(x)
RIGHT13(4(x)) → RIGHT13(x)
RIGHT13(5(x)) → RIGHT13(x)
RIGHT13(2(x)) → RIGHT13(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT13(1(x)) → RIGHT13(x)
The graph contains the following edges 1 > 1
- RIGHT13(0(x)) → RIGHT13(x)
The graph contains the following edges 1 > 1
- RIGHT13(4(x)) → RIGHT13(x)
The graph contains the following edges 1 > 1
- RIGHT13(5(x)) → RIGHT13(x)
The graph contains the following edges 1 > 1
- RIGHT13(2(x)) → RIGHT13(x)
The graph contains the following edges 1 > 1
(16) YES
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT12(1(x)) → RIGHT12(x)
RIGHT12(0(x)) → RIGHT12(x)
RIGHT12(4(x)) → RIGHT12(x)
RIGHT12(5(x)) → RIGHT12(x)
RIGHT12(2(x)) → RIGHT12(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT12(1(x)) → RIGHT12(x)
RIGHT12(0(x)) → RIGHT12(x)
RIGHT12(4(x)) → RIGHT12(x)
RIGHT12(5(x)) → RIGHT12(x)
RIGHT12(2(x)) → RIGHT12(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT12(1(x)) → RIGHT12(x)
The graph contains the following edges 1 > 1
- RIGHT12(0(x)) → RIGHT12(x)
The graph contains the following edges 1 > 1
- RIGHT12(4(x)) → RIGHT12(x)
The graph contains the following edges 1 > 1
- RIGHT12(5(x)) → RIGHT12(x)
The graph contains the following edges 1 > 1
- RIGHT12(2(x)) → RIGHT12(x)
The graph contains the following edges 1 > 1
(21) YES
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT11(1(x)) → RIGHT11(x)
RIGHT11(0(x)) → RIGHT11(x)
RIGHT11(4(x)) → RIGHT11(x)
RIGHT11(5(x)) → RIGHT11(x)
RIGHT11(2(x)) → RIGHT11(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT11(1(x)) → RIGHT11(x)
RIGHT11(0(x)) → RIGHT11(x)
RIGHT11(4(x)) → RIGHT11(x)
RIGHT11(5(x)) → RIGHT11(x)
RIGHT11(2(x)) → RIGHT11(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT11(1(x)) → RIGHT11(x)
The graph contains the following edges 1 > 1
- RIGHT11(0(x)) → RIGHT11(x)
The graph contains the following edges 1 > 1
- RIGHT11(4(x)) → RIGHT11(x)
The graph contains the following edges 1 > 1
- RIGHT11(5(x)) → RIGHT11(x)
The graph contains the following edges 1 > 1
- RIGHT11(2(x)) → RIGHT11(x)
The graph contains the following edges 1 > 1
(26) YES
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT10(1(x)) → RIGHT10(x)
RIGHT10(0(x)) → RIGHT10(x)
RIGHT10(4(x)) → RIGHT10(x)
RIGHT10(5(x)) → RIGHT10(x)
RIGHT10(2(x)) → RIGHT10(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(28) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT10(1(x)) → RIGHT10(x)
RIGHT10(0(x)) → RIGHT10(x)
RIGHT10(4(x)) → RIGHT10(x)
RIGHT10(5(x)) → RIGHT10(x)
RIGHT10(2(x)) → RIGHT10(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT10(1(x)) → RIGHT10(x)
The graph contains the following edges 1 > 1
- RIGHT10(0(x)) → RIGHT10(x)
The graph contains the following edges 1 > 1
- RIGHT10(4(x)) → RIGHT10(x)
The graph contains the following edges 1 > 1
- RIGHT10(5(x)) → RIGHT10(x)
The graph contains the following edges 1 > 1
- RIGHT10(2(x)) → RIGHT10(x)
The graph contains the following edges 1 > 1
(31) YES
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT9(1(x)) → RIGHT9(x)
RIGHT9(0(x)) → RIGHT9(x)
RIGHT9(4(x)) → RIGHT9(x)
RIGHT9(5(x)) → RIGHT9(x)
RIGHT9(2(x)) → RIGHT9(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(34) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT9(1(x)) → RIGHT9(x)
RIGHT9(0(x)) → RIGHT9(x)
RIGHT9(4(x)) → RIGHT9(x)
RIGHT9(5(x)) → RIGHT9(x)
RIGHT9(2(x)) → RIGHT9(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(35) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT9(1(x)) → RIGHT9(x)
The graph contains the following edges 1 > 1
- RIGHT9(0(x)) → RIGHT9(x)
The graph contains the following edges 1 > 1
- RIGHT9(4(x)) → RIGHT9(x)
The graph contains the following edges 1 > 1
- RIGHT9(5(x)) → RIGHT9(x)
The graph contains the following edges 1 > 1
- RIGHT9(2(x)) → RIGHT9(x)
The graph contains the following edges 1 > 1
(36) YES
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT8(1(x)) → RIGHT8(x)
RIGHT8(0(x)) → RIGHT8(x)
RIGHT8(4(x)) → RIGHT8(x)
RIGHT8(5(x)) → RIGHT8(x)
RIGHT8(2(x)) → RIGHT8(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(39) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT8(1(x)) → RIGHT8(x)
RIGHT8(0(x)) → RIGHT8(x)
RIGHT8(4(x)) → RIGHT8(x)
RIGHT8(5(x)) → RIGHT8(x)
RIGHT8(2(x)) → RIGHT8(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(40) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT8(1(x)) → RIGHT8(x)
The graph contains the following edges 1 > 1
- RIGHT8(0(x)) → RIGHT8(x)
The graph contains the following edges 1 > 1
- RIGHT8(4(x)) → RIGHT8(x)
The graph contains the following edges 1 > 1
- RIGHT8(5(x)) → RIGHT8(x)
The graph contains the following edges 1 > 1
- RIGHT8(2(x)) → RIGHT8(x)
The graph contains the following edges 1 > 1
(41) YES
(42) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT7(1(x)) → RIGHT7(x)
RIGHT7(0(x)) → RIGHT7(x)
RIGHT7(4(x)) → RIGHT7(x)
RIGHT7(5(x)) → RIGHT7(x)
RIGHT7(2(x)) → RIGHT7(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(43) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(44) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT7(1(x)) → RIGHT7(x)
RIGHT7(0(x)) → RIGHT7(x)
RIGHT7(4(x)) → RIGHT7(x)
RIGHT7(5(x)) → RIGHT7(x)
RIGHT7(2(x)) → RIGHT7(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(45) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT7(1(x)) → RIGHT7(x)
The graph contains the following edges 1 > 1
- RIGHT7(0(x)) → RIGHT7(x)
The graph contains the following edges 1 > 1
- RIGHT7(4(x)) → RIGHT7(x)
The graph contains the following edges 1 > 1
- RIGHT7(5(x)) → RIGHT7(x)
The graph contains the following edges 1 > 1
- RIGHT7(2(x)) → RIGHT7(x)
The graph contains the following edges 1 > 1
(46) YES
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)
RIGHT6(4(x)) → RIGHT6(x)
RIGHT6(5(x)) → RIGHT6(x)
RIGHT6(2(x)) → RIGHT6(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(48) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(49) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT6(1(x)) → RIGHT6(x)
RIGHT6(0(x)) → RIGHT6(x)
RIGHT6(4(x)) → RIGHT6(x)
RIGHT6(5(x)) → RIGHT6(x)
RIGHT6(2(x)) → RIGHT6(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(50) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT6(1(x)) → RIGHT6(x)
The graph contains the following edges 1 > 1
- RIGHT6(0(x)) → RIGHT6(x)
The graph contains the following edges 1 > 1
- RIGHT6(4(x)) → RIGHT6(x)
The graph contains the following edges 1 > 1
- RIGHT6(5(x)) → RIGHT6(x)
The graph contains the following edges 1 > 1
- RIGHT6(2(x)) → RIGHT6(x)
The graph contains the following edges 1 > 1
(51) YES
(52) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)
RIGHT5(4(x)) → RIGHT5(x)
RIGHT5(5(x)) → RIGHT5(x)
RIGHT5(2(x)) → RIGHT5(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(53) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(54) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT5(1(x)) → RIGHT5(x)
RIGHT5(0(x)) → RIGHT5(x)
RIGHT5(4(x)) → RIGHT5(x)
RIGHT5(5(x)) → RIGHT5(x)
RIGHT5(2(x)) → RIGHT5(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(55) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT5(1(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(0(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(4(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(5(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(2(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
(56) YES
(57) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(x)
RIGHT4(4(x)) → RIGHT4(x)
RIGHT4(5(x)) → RIGHT4(x)
RIGHT4(2(x)) → RIGHT4(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(58) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(59) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT4(1(x)) → RIGHT4(x)
RIGHT4(0(x)) → RIGHT4(x)
RIGHT4(4(x)) → RIGHT4(x)
RIGHT4(5(x)) → RIGHT4(x)
RIGHT4(2(x)) → RIGHT4(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(60) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT4(1(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(0(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(4(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(5(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(2(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
(61) YES
(62) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)
RIGHT3(4(x)) → RIGHT3(x)
RIGHT3(5(x)) → RIGHT3(x)
RIGHT3(2(x)) → RIGHT3(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(63) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(64) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT3(1(x)) → RIGHT3(x)
RIGHT3(0(x)) → RIGHT3(x)
RIGHT3(4(x)) → RIGHT3(x)
RIGHT3(5(x)) → RIGHT3(x)
RIGHT3(2(x)) → RIGHT3(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(65) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT3(1(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(0(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(4(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(5(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(2(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
(66) YES
(67) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)
RIGHT2(4(x)) → RIGHT2(x)
RIGHT2(5(x)) → RIGHT2(x)
RIGHT2(2(x)) → RIGHT2(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(68) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(69) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT2(1(x)) → RIGHT2(x)
RIGHT2(0(x)) → RIGHT2(x)
RIGHT2(4(x)) → RIGHT2(x)
RIGHT2(5(x)) → RIGHT2(x)
RIGHT2(2(x)) → RIGHT2(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(70) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT2(1(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(0(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(4(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(5(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(2(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
(71) YES
(72) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)
RIGHT1(4(x)) → RIGHT1(x)
RIGHT1(5(x)) → RIGHT1(x)
RIGHT1(2(x)) → RIGHT1(x)
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(73) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(74) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT1(1(x)) → RIGHT1(x)
RIGHT1(0(x)) → RIGHT1(x)
RIGHT1(4(x)) → RIGHT1(x)
RIGHT1(5(x)) → RIGHT1(x)
RIGHT1(2(x)) → RIGHT1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(75) QDPSizeChangeProof (EQUIVALENT transformation)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.
From the DPs we obtained the following set of size-change graphs:
- RIGHT1(1(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(0(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(4(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(5(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(2(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
(76) YES
(77) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Begin(5(4(5(x)))) → Wait(Right1(x))
Begin(4(5(x))) → Wait(Right2(x))
Begin(5(x)) → Wait(Right3(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right2(0(x)) → A0(Right2(x))
Right3(0(x)) → A0(Right3(x))
Right4(0(x)) → A0(Right4(x))
Right5(0(x)) → A0(Right5(x))
Right6(0(x)) → A0(Right6(x))
Right7(0(x)) → A0(Right7(x))
Right8(0(x)) → A0(Right8(x))
Right9(0(x)) → A0(Right9(x))
Right10(0(x)) → A0(Right10(x))
Right11(0(x)) → A0(Right11(x))
Right12(0(x)) → A0(Right12(x))
Right13(0(x)) → A0(Right13(x))
Right14(0(x)) → A0(Right14(x))
Right1(1(x)) → A1(Right1(x))
Right2(1(x)) → A1(Right2(x))
Right3(1(x)) → A1(Right3(x))
Right4(1(x)) → A1(Right4(x))
Right5(1(x)) → A1(Right5(x))
Right6(1(x)) → A1(Right6(x))
Right7(1(x)) → A1(Right7(x))
Right8(1(x)) → A1(Right8(x))
Right9(1(x)) → A1(Right9(x))
Right10(1(x)) → A1(Right10(x))
Right11(1(x)) → A1(Right11(x))
Right12(1(x)) → A1(Right12(x))
Right13(1(x)) → A1(Right13(x))
Right14(1(x)) → A1(Right14(x))
Right1(4(x)) → A4(Right1(x))
Right2(4(x)) → A4(Right2(x))
Right3(4(x)) → A4(Right3(x))
Right4(4(x)) → A4(Right4(x))
Right5(4(x)) → A4(Right5(x))
Right6(4(x)) → A4(Right6(x))
Right7(4(x)) → A4(Right7(x))
Right8(4(x)) → A4(Right8(x))
Right9(4(x)) → A4(Right9(x))
Right10(4(x)) → A4(Right10(x))
Right11(4(x)) → A4(Right11(x))
Right12(4(x)) → A4(Right12(x))
Right13(4(x)) → A4(Right13(x))
Right14(4(x)) → A4(Right14(x))
Right1(5(x)) → A5(Right1(x))
Right2(5(x)) → A5(Right2(x))
Right3(5(x)) → A5(Right3(x))
Right4(5(x)) → A5(Right4(x))
Right5(5(x)) → A5(Right5(x))
Right6(5(x)) → A5(Right6(x))
Right7(5(x)) → A5(Right7(x))
Right8(5(x)) → A5(Right8(x))
Right9(5(x)) → A5(Right9(x))
Right10(5(x)) → A5(Right10(x))
Right11(5(x)) → A5(Right11(x))
Right12(5(x)) → A5(Right12(x))
Right13(5(x)) → A5(Right13(x))
Right14(5(x)) → A5(Right14(x))
Right1(2(x)) → A2(Right1(x))
Right2(2(x)) → A2(Right2(x))
Right3(2(x)) → A2(Right3(x))
Right4(2(x)) → A2(Right4(x))
Right5(2(x)) → A2(Right5(x))
Right6(2(x)) → A2(Right6(x))
Right7(2(x)) → A2(Right7(x))
Right8(2(x)) → A2(Right8(x))
Right9(2(x)) → A2(Right9(x))
Right10(2(x)) → A2(Right10(x))
Right11(2(x)) → A2(Right11(x))
Right12(2(x)) → A2(Right12(x))
Right13(2(x)) → A2(Right13(x))
Right14(2(x)) → A2(Right14(x))
A0(Left(x)) → Left(0(x))
A1(Left(x)) → Left(1(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
Wait(Left(x)) → Begin(x)
4(5(4(5(x)))) → 4(4(5(5(x))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(78) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(79) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(0(x)) → A0(Right2(x))
Right2(1(x)) → A1(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right2(2(x)) → A2(Right2(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right1(1(x)) → A1(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Right1(2(x)) → A2(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(80) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
Right2(1(x)) → A1(Right2(x))
Right1(1(x)) → A1(Right1(x))
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(1(x1)) = 2 + 2·x1
POL(2(x1)) = 2·x1
POL(4(x1)) = 2·x1
POL(5(x1)) = x1
POL(A0(x1)) = x1
POL(A1(x1)) = 2 + 2·x1
POL(A2(x1)) = 2·x1
POL(A4(x1)) = 2·x1
POL(A5(x1)) = x1
POL(BEGIN(x1)) = x1
POL(End(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right1(x1)) = 2·x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = x1
POL(WAIT(x1)) = x1
(81) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(0(x)) → A0(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right2(2(x)) → A2(Right2(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Right1(2(x)) → A2(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(82) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
Right2(2(x)) → A2(Right2(x))
Right1(2(x)) → A2(Right1(x))
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = x1
POL(1(x1)) = x1
POL(2(x1)) = 2 + x1
POL(4(x1)) = 2·x1
POL(5(x1)) = x1
POL(A0(x1)) = x1
POL(A1(x1)) = x1
POL(A2(x1)) = 2 + x1
POL(A4(x1)) = 2·x1
POL(A5(x1)) = x1
POL(BEGIN(x1)) = 2·x1
POL(End(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right1(x1)) = 2·x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = x1
POL(WAIT(x1)) = 2·x1
(83) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(0(x)) → A0(Right2(x))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(0(x)) → A0(Right1(x))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(84) MRRProof (EQUIVALENT transformation)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented rules of the TRS R:
Right2(0(x)) → A0(Right2(x))
Right1(0(x)) → A0(Right1(x))
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = 1 + x1
POL(1(x1)) = x1
POL(2(x1)) = x1
POL(4(x1)) = 2·x1
POL(5(x1)) = x1
POL(A0(x1)) = 1 + x1
POL(A1(x1)) = x1
POL(A2(x1)) = x1
POL(A4(x1)) = 2·x1
POL(A5(x1)) = x1
POL(BEGIN(x1)) = x1
POL(End(x1)) = 2·x1
POL(Left(x1)) = x1
POL(Right1(x1)) = 2·x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = x1
POL(WAIT(x1)) = x1
(85) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(86) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
BEGIN(5(4(5(x)))) → WAIT(Right1(x))
BEGIN(4(5(x))) → WAIT(Right2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( A1(x1) ) = max{0, -2} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
(87) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Right2(4(5(End(x)))) → Left(4(4(5(5(End(x))))))
Right2(4(x)) → A4(Right2(x))
Right2(5(x)) → A5(Right2(x))
Right1(4(End(x))) → Left(4(4(5(5(End(x))))))
Right1(4(x)) → A4(Right1(x))
Right1(5(x)) → A5(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(88) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(89) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(90) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
WAIT(Left(x)) → BEGIN(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(0(x1)) = 1 + x1
POL(1(x1)) = 0
POL(2(x1)) = x1
POL(4(x1)) = 0
POL(5(x1)) = 1 + x1
POL(A0(x1)) = 1 + x1
POL(A1(x1)) = 1
POL(A2(x1)) = x1
POL(A4(x1)) = 1
POL(A5(x1)) = 1 + x1
POL(BEGIN(x1)) = 1 + x1
POL(End(x1)) = x1
POL(Left(x1)) = 1 + x1
POL(Right3(x1)) = 1 + x1
POL(WAIT(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
A4(Left(x)) → Left(4(x))
A5(Left(x)) → Left(5(x))
A2(Left(x)) → Left(2(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
(91) Obligation:
Q DP problem:
The TRS P consists of the following rules:
BEGIN(5(x)) → WAIT(Right3(x))
The TRS R consists of the following rules:
Right3(4(5(4(End(x))))) → Left(4(4(5(5(End(x))))))
Right3(0(x)) → A0(Right3(x))
Right3(1(x)) → A1(Right3(x))
Right3(4(x)) → A4(Right3(x))
Right3(5(x)) → A5(Right3(x))
Right3(2(x)) → A2(Right3(x))
A2(Left(x)) → Left(2(x))
A5(Left(x)) → Left(5(x))
A4(Left(x)) → Left(4(x))
4(5(4(5(x)))) → 4(4(5(5(x))))
A1(Left(x)) → Left(1(x))
A0(Left(x)) → Left(0(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(92) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
(93) TRUE