(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
a(c(x)) → c(a(b(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:
A(b(x)) → B(c(a(x)))
A(b(x)) → A(x)
B(c(x)) → B(b(x))
B(c(x)) → B(x)
A(c(x)) → A(b(x))
A(c(x)) → B(x)
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
a(c(x)) → c(a(b(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 2 SCCs with 2 less nodes.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → B(x)
B(c(x)) → B(b(x))
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
a(c(x)) → c(a(b(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(x)) → B(x)
B(c(x)) → B(b(x))
The TRS R consists of the following rules:
b(c(x)) → c(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(c(x)) → B(x)
B(c(x)) → B(b(x))
The TRS R consists of the following rules:
b(c(x)) → c(b(b(x)))
The set Q consists of the following terms:
b(c(x0))
We have to consider all minimal (P,Q,R)-chains.
(10) 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(x)) → B(x)
B(c(x)) → B(b(x))
Used ordering: Polynomial interpretation [POLO]:
POL(B(x1)) = 3·x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
(11) Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:
b(c(x)) → c(b(b(x)))
The set Q consists of the following terms:
b(c(x0))
We have to consider all minimal (P,Q,R)-chains.
(12) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(13) YES
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(x)) → A(b(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(c(a(x)))
b(c(x)) → c(b(b(x)))
a(c(x)) → c(a(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(15) 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.
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(c(x)) → A(b(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
b(c(x)) → c(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(17) 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:
A(c(x)) → A(b(x))
Used ordering: Polynomial interpretation [POLO]:
POL(A(x1)) = 3·x1
POL(b(x1)) = x1
POL(c(x1)) = 1 + x1
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
The TRS R consists of the following rules:
b(c(x)) → c(b(b(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(19) 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.
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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:
- A(b(x)) → A(x)
The graph contains the following edges 1 > 1
(22) YES