Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, x)
F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(f(x, y), z)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
The TRS R consists of the following rules:
f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, x)
F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(f(x, y), z)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
The TRS R consists of the following rules:
f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(x, f(y, z)) → F(x, y)
F(f(b, a), x) → F(b, x)
F(f(a, b), x) → F(a, x)
F(x, f(y, z)) → F(f(x, y), z)
F(f(b, a), x) → F(b, f(b, x))
F(f(a, b), x) → F(a, f(a, x))
The TRS R consists of the following rules:
f(f(a, b), x) → f(a, f(a, x))
f(f(b, a), x) → f(b, f(b, x))
f(x, f(y, z)) → f(f(x, y), z)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.