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:
f3(x, y, z) -> g3(x, y, z)
g3(0, 1, x) -> f3(x, x, x)
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(x, y, z) -> g3(x, y, z)
g3(0, 1, x) -> f3(x, x, x)
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(x, y, z) -> g3(x, y, z)
g3(0, 1, x) -> f3(x, x, x)
The set Q consists of the following terms:
f3(x0, x1, x2)
g3(0, 1, x0)
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G3(0, 1, x) -> F3(x, x, x)
F3(x, y, z) -> G3(x, y, z)
The TRS R consists of the following rules:
f3(x, y, z) -> g3(x, y, z)
g3(0, 1, x) -> f3(x, x, x)
The set Q consists of the following terms:
f3(x0, x1, x2)
g3(0, 1, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G3(0, 1, x) -> F3(x, x, x)
F3(x, y, z) -> G3(x, y, z)
The TRS R consists of the following rules:
f3(x, y, z) -> g3(x, y, z)
g3(0, 1, x) -> f3(x, x, x)
The set Q consists of the following terms:
f3(x0, x1, x2)
g3(0, 1, x0)
We have to consider all minimal (P,Q,R)-chains.