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(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(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:
G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
F1(g1(a)) -> F1(s1(g1(b)))
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
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
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
G1(x) -> F1(g1(x))
F1(g1(a)) -> G1(b)
G1(x) -> G1(x)
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F1(g1(a)) -> G1(b)
Used argument filtering: G1(x1) = x1
F1(x1) = x1
g1(x1) = x1
a = a
b = b
f1(x1) = f
Used ordering: Quasi Precedence:
a > [b, f]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G1(x) -> F1(g1(x))
G1(x) -> G1(x)
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
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
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(x) -> G1(x)
The TRS R consists of the following rules:
f1(g1(a)) -> f1(s1(g1(b)))
f1(f1(x)) -> b
g1(x) -> f1(g1(x))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.