Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F1(1) -> F1(g1(1))
G1(0) -> G1(f1(0))
F1(1) -> G1(1)
G1(0) -> F1(0)
The TRS R consists of the following rules:
f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)
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(1) -> F1(g1(1))
G1(0) -> G1(f1(0))
F1(1) -> G1(1)
G1(0) -> F1(0)
The TRS R consists of the following rules:
f1(1) -> f1(g1(1))
f1(f1(x)) -> f1(x)
g1(0) -> g1(f1(0))
g1(g1(x)) -> g1(x)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 4 less nodes.