(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
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(x, a(b(y))) → F(a(b(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
The TRS R consists of the following rules:
f(x, a(b(y))) → a(f(a(b(x)), y))
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) 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.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(y))) → F(a(b(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(5) QDPToSRSProof (SOUND transformation)
The finiteness of this DP problem is implied by strong termination of a SRS due to [UNKNOWN].
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → b(a(x))
Q is empty.
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) 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:
a(b(x)) → b(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(9) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(10) 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:
a(b(x)) → b(a(x))
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(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.
(13) 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))
(14) 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.
(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:
- A(b(x)) → A(x)
The graph contains the following edges 1 > 1
(16) TRUE