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(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
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(f(y, z), f(x, f(a, x))) → F(x, a)
F(f(y, z), f(x, f(a, x))) → F(a, z)
F(f(y, z), f(x, f(a, x))) → F(f(f(a, z), f(x, a)), f(a, y))
F(f(y, z), f(x, f(a, x))) → F(a, y)
F(f(y, z), f(x, f(a, x))) → F(f(a, z), f(x, a))
The TRS R consists of the following rules:
f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
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(f(y, z), f(x, f(a, x))) → F(x, a)
F(f(y, z), f(x, f(a, x))) → F(a, z)
F(f(y, z), f(x, f(a, x))) → F(f(f(a, z), f(x, a)), f(a, y))
F(f(y, z), f(x, f(a, x))) → F(a, y)
F(f(y, z), f(x, f(a, x))) → F(f(a, z), f(x, a))
The TRS R consists of the following rules:
f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
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
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F(f(y, z), f(x, f(a, x))) → F(x, a)
F(f(y, z), f(x, f(a, x))) → F(a, z)
F(f(y, z), f(x, f(a, x))) → F(f(f(a, z), f(x, a)), f(a, y))
F(f(y, z), f(x, f(a, x))) → F(a, y)
F(f(y, z), f(x, f(a, x))) → F(f(a, z), f(x, a))
The TRS R consists of the following rules:
f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 4 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(f(y, z), f(x, f(a, x))) → F(f(f(a, z), f(x, a)), f(a, y))
The TRS R consists of the following rules:
f(f(y, z), f(x, f(a, x))) → f(f(f(a, z), f(x, a)), f(a, y))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.