(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
BEGIN(c(a(b(b(c(x)))))) → RIGHT1(x)
BEGIN(a(b(b(c(x))))) → WAIT(Right2(x))
BEGIN(a(b(b(c(x))))) → RIGHT2(x)
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(b(c(x)))) → RIGHT3(x)
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(b(c(x))) → RIGHT4(x)
BEGIN(c(x)) → WAIT(Right5(x))
BEGIN(c(x)) → RIGHT5(x)
RIGHT1(b(End(x))) → B(c(b(b(c(a(End(x)))))))
RIGHT1(b(End(x))) → B(b(c(a(End(x)))))
RIGHT1(b(End(x))) → B(c(a(End(x))))
RIGHT2(b(c(End(x)))) → B(c(b(b(c(a(End(x)))))))
RIGHT2(b(c(End(x)))) → B(b(c(a(End(x)))))
RIGHT2(b(c(End(x)))) → B(c(a(End(x))))
RIGHT3(b(c(a(End(x))))) → B(c(b(b(c(a(End(x)))))))
RIGHT3(b(c(a(End(x))))) → B(b(c(a(End(x)))))
RIGHT4(b(c(a(b(End(x)))))) → B(c(b(b(c(a(End(x)))))))
RIGHT4(b(c(a(b(End(x)))))) → B(b(c(a(End(x)))))
RIGHT4(b(c(a(b(End(x)))))) → B(c(a(End(x))))
RIGHT5(b(c(a(b(b(End(x))))))) → B(c(b(b(c(a(End(x)))))))
RIGHT5(b(c(a(b(b(End(x))))))) → B(b(c(a(End(x)))))
RIGHT5(b(c(a(b(b(End(x))))))) → B(c(a(End(x))))
RIGHT1(b(x)) → AB(Right1(x))
RIGHT1(b(x)) → RIGHT1(x)
RIGHT2(b(x)) → AB(Right2(x))
RIGHT2(b(x)) → RIGHT2(x)
RIGHT3(b(x)) → AB(Right3(x))
RIGHT3(b(x)) → RIGHT3(x)
RIGHT4(b(x)) → AB(Right4(x))
RIGHT4(b(x)) → RIGHT4(x)
RIGHT5(b(x)) → AB(Right5(x))
RIGHT5(b(x)) → RIGHT5(x)
RIGHT1(c(x)) → AC(Right1(x))
RIGHT1(c(x)) → RIGHT1(x)
RIGHT2(c(x)) → AC(Right2(x))
RIGHT2(c(x)) → RIGHT2(x)
RIGHT3(c(x)) → AC(Right3(x))
RIGHT3(c(x)) → RIGHT3(x)
RIGHT4(c(x)) → AC(Right4(x))
RIGHT4(c(x)) → RIGHT4(x)
RIGHT5(c(x)) → AC(Right5(x))
RIGHT5(c(x)) → RIGHT5(x)
RIGHT1(a(x)) → AA(Right1(x))
RIGHT1(a(x)) → RIGHT1(x)
RIGHT2(a(x)) → AA(Right2(x))
RIGHT2(a(x)) → RIGHT2(x)
RIGHT3(a(x)) → AA(Right3(x))
RIGHT3(a(x)) → RIGHT3(x)
RIGHT4(a(x)) → AA(Right4(x))
RIGHT4(a(x)) → RIGHT4(x)
RIGHT5(a(x)) → AA(Right5(x))
RIGHT5(a(x)) → RIGHT5(x)
AB(Left(x)) → B(x)
WAIT(Left(x)) → BEGIN(x)
B(c(a(b(b(c(x)))))) → B(c(b(b(c(a(x))))))
B(c(a(b(b(c(x)))))) → B(b(c(a(x))))
B(c(a(b(b(c(x)))))) → B(c(a(x)))
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 7 SCCs with 37 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(a(b(b(c(x)))))) → B(c(a(x)))
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) 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.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(a(b(b(c(x)))))) → B(c(a(x)))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) 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 dependency pairs:
B(c(a(b(b(c(x)))))) → B(c(a(x)))
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = 2·x1
POL(a(x1)) = 2·x1
POL(b(x1)) = 2 + 2·x1
POL(c(x1)) = 2·x1
(9) Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(11) YES
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(x)
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
RIGHT5(c(x)) → RIGHT5(x)
RIGHT5(b(x)) → RIGHT5(x)
RIGHT5(a(x)) → RIGHT5(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:
- RIGHT5(c(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(b(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
- RIGHT5(a(x)) → RIGHT5(x)
The graph contains the following edges 1 > 1
(16) YES
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(x)
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
RIGHT4(c(x)) → RIGHT4(x)
RIGHT4(b(x)) → RIGHT4(x)
RIGHT4(a(x)) → RIGHT4(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:
- RIGHT4(c(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(b(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
- RIGHT4(a(x)) → RIGHT4(x)
The graph contains the following edges 1 > 1
(21) YES
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(x)
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
RIGHT3(c(x)) → RIGHT3(x)
RIGHT3(b(x)) → RIGHT3(x)
RIGHT3(a(x)) → RIGHT3(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:
- RIGHT3(c(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(b(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
- RIGHT3(a(x)) → RIGHT3(x)
The graph contains the following edges 1 > 1
(26) YES
(27) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(x)
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
RIGHT2(c(x)) → RIGHT2(x)
RIGHT2(b(x)) → RIGHT2(x)
RIGHT2(a(x)) → RIGHT2(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:
- RIGHT2(c(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(b(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
- RIGHT2(a(x)) → RIGHT2(x)
The graph contains the following edges 1 > 1
(31) YES
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(x)
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
RIGHT1(c(x)) → RIGHT1(x)
RIGHT1(b(x)) → RIGHT1(x)
RIGHT1(a(x)) → RIGHT1(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:
- RIGHT1(c(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(b(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
- RIGHT1(a(x)) → RIGHT1(x)
The graph contains the following edges 1 > 1
(36) YES
(37) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
BEGIN(a(b(b(c(x))))) → WAIT(Right2(x))
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Begin(c(a(b(b(c(x)))))) → Wait(Right1(x))
Begin(a(b(b(c(x))))) → Wait(Right2(x))
Begin(b(b(c(x)))) → Wait(Right3(x))
Begin(b(c(x))) → Wait(Right4(x))
Begin(c(x)) → Wait(Right5(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(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(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))
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))
Ab(Left(x)) → Left(b(x))
Ac(Left(x)) → Left(c(x))
Aa(Left(x)) → Left(a(x))
Wait(Left(x)) → Begin(x)
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(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:
WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
BEGIN(a(b(b(c(x))))) → WAIT(Right2(x))
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(40) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
BEGIN(a(b(b(c(x))))) → WAIT(Right2(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(Left(x1)) = | | + | | / | 0A | -I | 0A | \ |
| | | -I | -I | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(BEGIN(x1)) = | -I | + | | · | x1 |
| POL(c(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 1A | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | -I | 0A | | |
| \ | -I | 0A | 0A | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | 1A | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(Right1(x1)) = | | + | | / | 1A | 0A | -I | \ |
| | | 0A | 0A | 1A | | |
| \ | 0A | 1A | 0A | / |
| · | x1 |
| POL(Right2(x1)) = | | + | | / | 0A | -I | 0A | \ |
| | | -I | -I | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(Right3(x1)) = | | + | | / | 1A | 0A | 0A | \ |
| | | 0A | 0A | 1A | | |
| \ | 0A | 1A | 0A | / |
| · | x1 |
| POL(Right4(x1)) = | | + | | / | 0A | -I | 0A | \ |
| | | -I | -I | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(Right5(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | -I | -I | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(End(x1)) = | | + | | / | 0A | -I | -I | \ |
| | | 1A | -I | 1A | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(Ab(x1)) = | | + | | / | 0A | 1A | 0A | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(Ac(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | -I | -I | 0A | | |
| \ | 0A | -I | 1A | / |
| · | x1 |
| POL(Aa(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | -I | 0A | 0A | | |
| \ | 0A | 0A | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
(41) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right2(b(c(End(x)))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right2(b(x)) → Ab(Right2(x))
Right2(c(x)) → Ac(Right2(x))
Right2(a(x)) → Aa(Right2(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(42) 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.
(43) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(44) QDPOrderProof (EQUIVALENT transformation)
We use the reduction pair processor [LPAR04,JAR06].
The following pairs can be oriented strictly and are deleted.
BEGIN(c(a(b(b(c(x)))))) → WAIT(Right1(x))
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO] with arctic natural numbers [ARCTIC]:
| POL(Left(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(BEGIN(x1)) = | 0A | + | | · | x1 |
| POL(c(x1)) = | | + | | / | -I | -I | -I | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(a(x1)) = | | + | | / | -I | -I | 0A | \ |
| | | 0A | 0A | 1A | | |
| \ | -I | -I | -I | / |
| · | x1 |
| POL(b(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | 0A | | |
| \ | 0A | -I | -I | / |
| · | x1 |
| POL(Right1(x1)) = | | + | | / | 0A | 0A | -I | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(Right3(x1)) = | | + | | / | -I | 0A | 0A | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(Right4(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(Right5(x1)) = | | + | | / | -I | 0A | -I | \ |
| | | 0A | -I | -I | | |
| \ | -I | -I | 0A | / |
| · | x1 |
| POL(End(x1)) = | | + | | / | -I | -I | -I | \ |
| | | -I | -I | -I | | |
| \ | -I | -I | -I | / |
| · | x1 |
| POL(Ab(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | 0A | 0A | 0A | | |
| \ | -I | 0A | -I | / |
| · | x1 |
| POL(Ac(x1)) = | | + | | / | 0A | 0A | 0A | \ |
| | | -I | -I | -I | | |
| \ | 0A | 0A | 0A | / |
| · | x1 |
| POL(Aa(x1)) = | | + | | / | 0A | 0A | 1A | \ |
| | | -I | -I | 0A | | |
| \ | -I | -I | -I | / |
| · | x1 |
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
(45) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right1(b(End(x))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right1(b(x)) → Ab(Right1(x))
Right1(c(x)) → Ac(Right1(x))
Right1(a(x)) → Aa(Right1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(46) 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.
(47) Obligation:
Q DP problem:
The TRS P consists of the following rules:
WAIT(Left(x)) → BEGIN(x)
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(48) 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(Aa(x1)) = 1
POL(Ab(x1)) = x1
POL(Ac(x1)) = 1 + x1
POL(BEGIN(x1)) = x1
POL(End(x1)) = x1
POL(Left(x1)) = 1 + x1
POL(Right3(x1)) = 1 + x1
POL(Right4(x1)) = 1 + x1
POL(Right5(x1)) = 1 + x1
POL(WAIT(x1)) = x1
POL(a(x1)) = 0
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
Aa(Left(x)) → Left(a(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
(49) Obligation:
Q DP problem:
The TRS P consists of the following rules:
BEGIN(b(b(c(x)))) → WAIT(Right3(x))
BEGIN(b(c(x))) → WAIT(Right4(x))
BEGIN(c(x)) → WAIT(Right5(x))
The TRS R consists of the following rules:
Right5(b(c(a(b(b(End(x))))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right5(b(x)) → Ab(Right5(x))
Right5(c(x)) → Ac(Right5(x))
Right5(a(x)) → Aa(Right5(x))
Aa(Left(x)) → Left(a(x))
Ac(Left(x)) → Left(c(x))
Ab(Left(x)) → Left(b(x))
b(c(a(b(b(c(x)))))) → a(b(c(b(b(c(a(x)))))))
Right4(b(c(a(b(End(x)))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right4(b(x)) → Ab(Right4(x))
Right4(c(x)) → Ac(Right4(x))
Right4(a(x)) → Aa(Right4(x))
Right3(b(c(a(End(x))))) → Left(a(b(c(b(b(c(a(End(x)))))))))
Right3(b(x)) → Ab(Right3(x))
Right3(c(x)) → Ac(Right3(x))
Right3(a(x)) → Aa(Right3(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(50) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 3 less nodes.
(51) TRUE