(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(x, a(b(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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(x1)))
f(x0, b(c(x1)))
f(x0, c(a(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(y))) → F(a(b(x)), y)
F(x, b(c(y))) → F(b(c(x)), y)
F(x, c(a(y))) → F(c(a(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(y))) → f(a(b(x)), y)
f(x, b(c(y))) → f(b(c(x)), y)
f(x, c(a(y))) → f(c(a(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(x1)))
f(x0, b(c(x1)))
f(x0, c(a(x1)))
f(a(x0), x1)
f(b(x0), x1)
f(c(x0), x1)
We have to consider all minimal (P,Q,R)-chains.