(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(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:
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, x)
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, f(b, x))
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(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.
(4) Complex Obligation (AND)
(5) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, f(b, x))
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(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:
F(b, f(a, x)) → F(b, f(b, x))
F(b, f(a, x)) → F(b, f(b, f(b, x)))
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(8) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(10) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(11) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
The TRS R consists of the following rules:
f(b, f(a, x)) → f(b, f(b, f(b, x)))
The set Q consists of the following terms:
f(b, f(a, x0))
We have to consider all minimal (P,Q,R)-chains.
(12) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(13) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(b, f(a, x)) → F(b, x)
R is empty.
The set Q consists of the following terms:
f(b, f(a, x0))
We have to consider all minimal (P,Q,R)-chains.
(14) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
b1(a(x)) → b1(x)
R is empty.
The set Q consists of the following terms:
b(a(x0))
We have to consider all minimal (P,Q,R)-chains.
(16) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
b(a(x0))
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
b1(a(x)) → b1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) 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:
- b1(a(x)) → b1(x)
The graph contains the following edges 1 > 1
(19) YES
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
f(b, f(a, x)) → f(b, f(b, f(b, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(21) 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.
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, f(a, x))
F(a, f(b, x)) → F(a, f(a, f(a, x)))
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(23) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(25) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
The TRS R consists of the following rules:
f(a, f(b, x)) → f(a, f(a, f(a, x)))
The set Q consists of the following terms:
f(a, f(b, x0))
We have to consider all minimal (P,Q,R)-chains.
(27) UsableRulesProof (EQUIVALENT transformation)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(a, f(b, x)) → F(a, x)
R is empty.
The set Q consists of the following terms:
f(a, f(b, x0))
We have to consider all minimal (P,Q,R)-chains.
(29) ATransformationProof (EQUIVALENT transformation)
We have applied the A-Transformation [FROCOS05] to get from an applicative problem to a standard problem.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → a1(x)
R is empty.
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(31) QReductionProof (EQUIVALENT transformation)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].
a(b(x0))
(32) Obligation:
Q DP problem:
The TRS P consists of the following rules:
a1(b(x)) → a1(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) 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:
- a1(b(x)) → a1(x)
The graph contains the following edges 1 > 1
(34) YES