(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(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(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The signature Sigma is {
f}
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The set Q consists of the following terms:
f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(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(c(y)))) → F(b(c(a(b(x)))), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(y))
The TRS R consists of the following rules:
f(x, a(b(c(y)))) → f(b(c(a(b(x)))), y)
f(a(x), y) → f(x, a(y))
f(b(x), y) → f(x, b(y))
f(c(x), y) → f(x, c(y))
The set Q consists of the following terms:
f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(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(c(y)))) → F(b(c(a(b(x)))), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(y))
R is empty.
The set Q consists of the following terms:
f(x0, a(b(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(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(c(x1))))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
F(x, a(b(c(y)))) → F(b(c(a(b(x)))), y)
F(a(x), y) → F(x, a(y))
F(b(x), y) → F(x, b(y))
F(c(x), y) → F(x, c(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(c(x))) → b(a(c(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(c(x))) → A(c(b(x)))
The TRS R consists of the following rules:
a(b(c(x))) → b(a(c(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 0 SCCs with 1 less node.
(14) TRUE