(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(a(x)), y)
f(x, b(a(y))) → f(b(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
Q is empty.
(1) AAECC Innermost (EQUIVALENT transformation)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none
The TRS R 2 is
f(x, a(b(y))) → f(a(a(x)), y)
f(x, b(a(y))) → f(b(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(a(x)), y)
f(x, b(a(y))) → f(b(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
(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:
F(x, a(b(y))) → F(a(a(x)), y)
F(x, b(a(y))) → F(b(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))) → f(a(a(x)), y)
f(x, b(a(y))) → f(b(b(x)), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(5) 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.
(6) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(y))) → F(a(a(x)), y)
F(x, b(a(y))) → F(b(b(x)), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
R is empty.
The set Q consists of the following terms:
f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
We have to consider all minimal (P,Q,R)-chains.
(7) 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].
f(x0, a(b(x1)))
f(x0, b(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(y))) → F(a(a(x)), y)
F(x, b(a(y))) → F(b(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.
(9) QDPToSRSProof (SOUND transformation)
The finiteness of this DP problem is implied by strong termination of a SRS due to [UNKNOWN].
(10) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
a(b(x)) → a(a(x))
b(a(x)) → b(b(x))
Q is empty.
(11) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(a(x))
A(b(x)) → A(x)
B(a(x)) → B(b(x))
B(a(x)) → B(x)
The TRS R consists of the following rules:
a(b(x)) → a(a(x))
b(a(x)) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs.
(14) Complex Obligation (AND)
(15) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(x)
B(a(x)) → B(b(x))
The TRS R consists of the following rules:
a(b(x)) → a(a(x))
b(a(x)) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) 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.
(17) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(x)
B(a(x)) → B(b(x))
The TRS R consists of the following rules:
b(a(x)) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(19) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(20) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(21) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(x)
The TRS R consists of the following rules:
b(a(x)) → b(b(x))
The set Q consists of the following terms:
b(a(x0))
We have to consider all minimal (P,Q,R)-chains.
(22) 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.
(23) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(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.
(24) 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))
(25) Obligation:
Q DP problem:
The TRS P consists of the following rules:
B(a(x)) → B(x)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(26) 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:
- B(a(x)) → B(x)
The graph contains the following edges 1 > 1
(27) TRUE
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
A(b(x)) → A(a(x))
The TRS R consists of the following rules:
a(b(x)) → a(a(x))
b(a(x)) → b(b(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) 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.
(30) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(x)
A(b(x)) → A(a(x))
The TRS R consists of the following rules:
a(b(x)) → a(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(31) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.
(32) 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)) → a(a(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(33) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(34) 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)) → a(a(x))
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(35) 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.
(36) 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.
(37) 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))
(38) 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.
(39) 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
(40) TRUE