(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(a(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(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(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(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(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(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(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(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(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(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(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)) → b(a(a(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)
The TRS R consists of the following rules:
a(b(x)) → b(a(a(x)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(13) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
A(b(x)) → A(a(x))
A(b(x)) → A(x)
The TRS R consists of the following rules:
a(b(x)) → b(a(a(x)))
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(15) 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(b(x)) → A(a(x))
A(b(x)) → A(x)
Strictly oriented rules of the TRS R:
a(b(x)) → b(a(a(x)))
Used ordering: Knuth-Bendix order [KBO] with precedence:
a1 > A1 > b1
and weight map:
a_1=0
b_1=1
A_1=1
The variable weight is 1
(16) Obligation:
Q DP problem:
P is empty.
R is empty.
The set Q consists of the following terms:
a(b(x0))
We have to consider all minimal (P,Q,R)-chains.
(17) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(18) YES