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(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
F3(g1(x), s1(0), y) -> G1(s1(0))
F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))
G1(s1(x)) -> G1(x)
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
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:
F3(g1(x), s1(0), y) -> G1(s1(0))
F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))
G1(s1(x)) -> G1(x)
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G1(s1(x)) -> G1(x)
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
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.
G1(s1(x)) -> G1(x)
Used argument filtering: G1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
F3(g1(x), s1(0), y) -> F3(g1(s1(0)), y, g1(x))
The TRS R consists of the following rules:
f3(g1(x), s1(0), y) -> f3(g1(s1(0)), y, g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.