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:
f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(g2(x, x))
F1(s1(x)) -> G2(x, x)
The TRS R consists of the following rules:
f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(g2(x, x))
F1(s1(x)) -> G2(x, x)
The TRS R consists of the following rules:
f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(g2(x, x))
The TRS R consists of the following rules:
f1(s1(x)) -> f1(g2(x, x))
g2(0, 1) -> s1(0)
0 -> 1
Q is empty.
We have to consider all minimal (P,Q,R)-chains.