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(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(f(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
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(s(x), 0), f(y, z)) → F(y, s(z))
F(f(s(x), s(y)), f(z, w)) → F(x, y)
F(f(s(x), s(y)), f(z, w)) → F(f(x, y), f(z, w))
F(f(s(x), 0), f(y, z)) → F(f(y, z), f(y, s(z)))
The TRS R consists of the following rules:
f(f(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
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(s(x), 0), f(y, z)) → F(y, s(z))
F(f(s(x), s(y)), f(z, w)) → F(x, y)
F(f(s(x), s(y)), f(z, w)) → F(f(x, y), f(z, w))
F(f(s(x), 0), f(y, z)) → F(f(y, z), f(y, s(z)))
The TRS R consists of the following rules:
f(f(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
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(s(x), 0), f(y, z)) → F(y, s(z))
F(f(s(x), 0), f(y, z)) → F(f(y, z), f(y, s(z)))
F(f(s(x), s(y)), f(z, w)) → F(f(x, y), f(z, w))
F(f(s(x), s(y)), f(z, w)) → F(x, y)
The TRS R consists of the following rules:
f(f(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F(f(s(x), s(y)), f(z, w)) → F(x, y)
F(f(s(x), 0), f(y, z)) → F(f(y, z), f(y, s(z)))
F(f(s(x), s(y)), f(z, w)) → F(f(x, y), f(z, w))
The TRS R consists of the following rules:
f(f(s(x), 0), f(y, z)) → f(f(y, z), f(y, s(z)))
f(f(s(x), s(y)), f(z, w)) → f(f(x, y), f(z, w))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.