(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Begin(a(b(c(c(b(x)))))) → Wait(Right1(x))
Begin(b(c(c(b(x))))) → Wait(Right2(x))
Begin(c(c(b(x)))) → Wait(Right3(x))
Begin(c(b(x))) → Wait(Right4(x))
Begin(b(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(b(c(c(b(a(End(x)))))))))
Right2(b(a(End(x)))) → Left(a(b(b(c(c(b(a(End(x)))))))))
Right3(b(a(b(End(x))))) → Left(a(b(b(c(c(b(a(End(x)))))))))
Right4(b(a(b(c(End(x)))))) → Left(a(b(b(c(c(b(a(End(x)))))))))
Right5(b(a(b(c(c(End(x))))))) → Left(a(b(b(c(c(b(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right2(b(x)) → Ab(Right2(x))
Right3(b(x)) → Ab(Right3(x))
Right4(b(x)) → Ab(Right4(x))
Right5(b(x)) → Ab(Right5(x))
Right1(a(x)) → Aa(Right1(x))
Right2(a(x)) → Aa(Right2(x))
Right3(a(x)) → Aa(Right3(x))
Right4(a(x)) → Aa(Right4(x))
Right5(a(x)) → Aa(Right5(x))
Right1(c(x)) → Ac(Right1(x))
Right2(c(x)) → Ac(Right2(x))
Right3(c(x)) → Ac(Right3(x))
Right4(c(x)) → Ac(Right4(x))
Right5(c(x)) → Ac(Right5(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Wait(Left(x)) → Begin(x)
b(a(b(c(c(b(x)))))) → a(b(b(c(c(b(a(x)))))))
Q is empty.
(1) QTRS Reverse (EQUIVALENT transformation)
We applied the QTRS Reverse Processor [REVERSE].
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
End(b(Right1(x))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(a(b(Right2(x)))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(b(a(b(Right3(x))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(b(a(b(Right4(x)))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(c(b(a(b(Right5(x))))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(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:
END(b(Right1(x))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(Right1(x))) → A(b(c(c(b(b(a(Left(x))))))))
END(b(Right1(x))) → B(c(c(b(b(a(Left(x)))))))
END(b(Right1(x))) → C(c(b(b(a(Left(x))))))
END(b(Right1(x))) → C(b(b(a(Left(x)))))
END(b(Right1(x))) → B(b(a(Left(x))))
END(b(Right1(x))) → B(a(Left(x)))
END(b(Right1(x))) → A(Left(x))
END(b(Right1(x))) → LEFT(x)
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(a(b(Right2(x)))) → A(b(c(c(b(b(a(Left(x))))))))
END(a(b(Right2(x)))) → B(c(c(b(b(a(Left(x)))))))
END(a(b(Right2(x)))) → C(c(b(b(a(Left(x))))))
END(a(b(Right2(x)))) → C(b(b(a(Left(x)))))
END(a(b(Right2(x)))) → B(b(a(Left(x))))
END(a(b(Right2(x)))) → B(a(Left(x)))
END(a(b(Right2(x)))) → A(Left(x))
END(a(b(Right2(x)))) → LEFT(x)
END(b(a(b(Right3(x))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(a(b(Right3(x))))) → A(b(c(c(b(b(a(Left(x))))))))
END(b(a(b(Right3(x))))) → B(c(c(b(b(a(Left(x)))))))
END(b(a(b(Right3(x))))) → C(c(b(b(a(Left(x))))))
END(b(a(b(Right3(x))))) → C(b(b(a(Left(x)))))
END(b(a(b(Right3(x))))) → B(b(a(Left(x))))
END(b(a(b(Right3(x))))) → B(a(Left(x)))
END(b(a(b(Right3(x))))) → A(Left(x))
END(b(a(b(Right3(x))))) → LEFT(x)
END(c(b(a(b(Right4(x)))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(b(a(b(Right4(x)))))) → A(b(c(c(b(b(a(Left(x))))))))
END(c(b(a(b(Right4(x)))))) → B(c(c(b(b(a(Left(x)))))))
END(c(b(a(b(Right4(x)))))) → C(c(b(b(a(Left(x))))))
END(c(b(a(b(Right4(x)))))) → C(b(b(a(Left(x)))))
END(c(b(a(b(Right4(x)))))) → B(b(a(Left(x))))
END(c(b(a(b(Right4(x)))))) → B(a(Left(x)))
END(c(b(a(b(Right4(x)))))) → A(Left(x))
END(c(b(a(b(Right4(x)))))) → LEFT(x)
END(c(c(b(a(b(Right5(x))))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(c(b(a(b(Right5(x))))))) → A(b(c(c(b(b(a(Left(x))))))))
END(c(c(b(a(b(Right5(x))))))) → B(c(c(b(b(a(Left(x)))))))
END(c(c(b(a(b(Right5(x))))))) → C(c(b(b(a(Left(x))))))
END(c(c(b(a(b(Right5(x))))))) → C(b(b(a(Left(x)))))
END(c(c(b(a(b(Right5(x))))))) → B(b(a(Left(x))))
END(c(c(b(a(b(Right5(x))))))) → B(a(Left(x)))
END(c(c(b(a(b(Right5(x))))))) → A(Left(x))
END(c(c(b(a(b(Right5(x))))))) → LEFT(x)
LEFT(Ab(x)) → B(Left(x))
LEFT(Ab(x)) → LEFT(x)
LEFT(Aa(x)) → A(Left(x))
LEFT(Aa(x)) → LEFT(x)
LEFT(Ac(x)) → C(Left(x))
LEFT(Ac(x)) → LEFT(x)
B(c(c(b(a(b(x)))))) → A(b(c(c(b(b(a(x)))))))
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
B(c(c(b(a(b(x)))))) → C(c(b(b(a(x)))))
B(c(c(b(a(b(x)))))) → C(b(b(a(x))))
B(c(c(b(a(b(x)))))) → B(b(a(x)))
B(c(c(b(a(b(x)))))) → B(a(x))
B(c(c(b(a(b(x)))))) → A(x)
The TRS R consists of the following rules:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
End(b(Right1(x))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(a(b(Right2(x)))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(b(a(b(Right3(x))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(b(a(b(Right4(x)))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(c(b(a(b(Right5(x))))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(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 3 SCCs with 47 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(b(a(x)))
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
B(c(c(b(a(b(x)))))) → B(a(x))
The TRS R consists of the following rules:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
End(b(Right1(x))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(a(b(Right2(x)))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(b(a(b(Right3(x))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(b(a(b(Right4(x)))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(c(b(a(b(Right5(x))))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(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:
B(c(c(b(a(b(x)))))) → B(b(a(x)))
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
B(c(c(b(a(b(x)))))) → B(a(x))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(12) 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:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = x1
POL(Ac(x1)) = x1
POL(B(x1)) = x1
POL(Begin(x1)) = 3 + 2·x1
POL(Right1(x1)) = 3 + 3·x1
POL(Right2(x1)) = 3 + 2·x1
POL(Right3(x1)) = 1 + 2·x1
POL(Right4(x1)) = x1
POL(Right5(x1)) = x1
POL(Wait(x1)) = 1 + 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 2·x1
POL(c(x1)) = 2·x1
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(14) 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:
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = x1
POL(Ac(x1)) = 1 + x1
POL(B(x1)) = 2·x1
POL(Right1(x1)) = x1
POL(Right2(x1)) = x1
POL(Right3(x1)) = x1
POL(Right4(x1)) = 3 + 2·x1
POL(Right5(x1)) = 2 + x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = 2 + 2·x1
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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:
b(Right4(x)) → Right4(Ab(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = 2·x1
POL(B(x1)) = 2·x1
POL(Right1(x1)) = x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = 2·x1
POL(Right4(x1)) = 1 + 2·x1
POL(Right5(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = 2·x1
POL(c(x1)) = 2·x1
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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:
b(Right5(x)) → Right5(Ab(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = x1
POL(B(x1)) = x1
POL(Right1(x1)) = x1
POL(Right2(x1)) = 2·x1
POL(Right3(x1)) = 2·x1
POL(Right4(x1)) = x1
POL(Right5(x1)) = 2·x1
POL(a(x1)) = x1
POL(b(x1)) = 1 + x1
POL(c(x1)) = 2·x1
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
B(c(c(b(a(b(x)))))) → B(c(c(b(b(a(x))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial Order [NEGPOLO,POLO] with Interpretation:
| POL( Right1(x1) ) = max{0, 2x1 - 2} |
| POL( Right2(x1) ) = max{0, x1 - 2} |
| POL( Right4(x1) ) = max{0, -2} |
| POL( b(x1) ) = max{0, x1 - 1} |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
(21) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(22) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(23) YES
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
The TRS R consists of the following rules:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
End(b(Right1(x))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(a(b(Right2(x)))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(b(a(b(Right3(x))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(b(a(b(Right4(x)))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(c(b(a(b(Right5(x))))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) 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.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LEFT(Aa(x)) → LEFT(x)
LEFT(Ab(x)) → LEFT(x)
LEFT(Ac(x)) → LEFT(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) 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:
- LEFT(Aa(x)) → LEFT(x)
The graph contains the following edges 1 > 1
- LEFT(Ab(x)) → LEFT(x)
The graph contains the following edges 1 > 1
- LEFT(Ac(x)) → LEFT(x)
The graph contains the following edges 1 > 1
(28) YES
(29) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(Right1(x))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(a(b(Right3(x))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(b(a(b(Right4(x)))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(c(b(a(b(Right5(x))))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The TRS R consists of the following rules:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
End(b(Right1(x))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(a(b(Right2(x)))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(b(a(b(Right3(x))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(b(a(b(Right4(x)))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
End(c(c(b(a(b(Right5(x))))))) → End(a(b(c(c(b(b(a(Left(x)))))))))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(30) 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.
(31) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(Right1(x))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(a(b(Right3(x))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(b(a(b(Right4(x)))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(c(b(a(b(Right5(x))))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The TRS R consists of the following rules:
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(32) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
END(b(Right1(x))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(b(a(b(Right3(x))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(b(a(b(Right4(x)))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
END(c(c(b(a(b(Right5(x))))))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = x1
POL(Ac(x1)) = x1
POL(Begin(x1)) = x1
POL(END(x1)) = x1
POL(Left(x1)) = 1
POL(Right1(x1)) = 0
POL(Right2(x1)) = 0
POL(Right3(x1)) = 0
POL(Right4(x1)) = 0
POL(Right5(x1)) = 0
POL(Wait(x1)) = x1
POL(a(x1)) = 0
POL(b(x1)) = 1
POL(c(x1)) = 1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
(33) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The TRS R consists of the following rules:
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(34) 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:
b(c(c(b(a(Begin(x)))))) → Right1(Wait(x))
b(c(c(Begin(x)))) → Right3(Wait(x))
b(c(Begin(x))) → Right4(Wait(x))
b(Begin(x)) → Right5(Wait(x))
Used ordering: Polynomial interpretation [POLO]:
POL(Aa(x1)) = x1
POL(Ab(x1)) = x1
POL(Ac(x1)) = x1
POL(Begin(x1)) = 2 + x1
POL(END(x1)) = 2·x1
POL(Left(x1)) = 2 + x1
POL(Right1(x1)) = 1 + x1
POL(Right2(x1)) = 2 + x1
POL(Right3(x1)) = x1
POL(Right4(x1)) = x1
POL(Right5(x1)) = x1
POL(Wait(x1)) = x1
POL(a(x1)) = x1
POL(b(x1)) = x1
POL(c(x1)) = x1
(35) Obligation:
Q DP problem:
The TRS P consists of the following rules:
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The TRS R consists of the following rules:
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(36) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
END(a(b(Right2(x)))) → END(a(b(c(c(b(b(a(Left(x)))))))))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(a(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | -I | 0A | -I | | |
| \ | 0A | 0A | -I | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | -I | 0A | / |
| · | x1 |
| POL(Right2(x1)) = | | + | | / | 1A | 0A | -I | \ |
| | | 0A | 1A | 0A | | |
| \ | 0A | 0A | 1A | / |
| · | x1 |
| POL(c(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 1A | 1A | 0A | / |
| · | x1 |
| POL(Left(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(Ab(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | -I | 0A | / |
| · | x1 |
| POL(Aa(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | -I | 0A | -I | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(Ac(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | 0A | -I | | |
| \ | 1A | 1A | 0A | / |
| · | x1 |
| POL(Wait(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | -I | -I | -I | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(Begin(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | -I | / |
| · | x1 |
| POL(Right1(x1)) = | | + | | / | 1A | 1A | 0A | \ |
| | | 0A | 0A | 0A | | |
| \ | -I | -I | 1A | / |
| · | x1 |
| POL(Right3(x1)) = | | + | | / | 1A | 1A | 0A | \ |
| | | 0A | 0A | -I | | |
| \ | 0A | 0A | 1A | / |
| · | x1 |
| POL(Right4(x1)) = | | + | | / | 1A | -I | 0A | \ |
| | | 0A | 1A | 0A | | |
| \ | 0A | 0A | 1A | / |
| · | x1 |
| POL(Right5(x1)) = | | + | | / | 1A | 0A | 0A | \ |
| | | -I | 1A | 0A | | |
| \ | 1A | 1A | 1A | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
(37) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
Left(Ab(x)) → b(Left(x))
Left(Aa(x)) → a(Left(x))
Left(Ac(x)) → c(Left(x))
Left(Wait(x)) → Begin(x)
a(Right1(x)) → Right1(Aa(x))
a(Right2(x)) → Right2(Aa(x))
a(Right3(x)) → Right3(Aa(x))
a(Right4(x)) → Right4(Aa(x))
a(Right5(x)) → Right5(Aa(x))
b(c(c(b(Begin(x))))) → Right2(Wait(x))
b(Right1(x)) → Right1(Ab(x))
b(Right2(x)) → Right2(Ab(x))
b(Right3(x)) → Right3(Ab(x))
b(Right4(x)) → Right4(Ab(x))
b(Right5(x)) → Right5(Ab(x))
b(c(c(b(a(b(x)))))) → a(b(c(c(b(b(a(x)))))))
c(Right1(x)) → Right1(Ac(x))
c(Right2(x)) → Right2(Ac(x))
c(Right3(x)) → Right3(Ac(x))
c(Right4(x)) → Right4(Ac(x))
c(Right5(x)) → Right5(Ac(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(38) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(39) YES