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:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → G(x)
F(g(x), s(0), y) → F(y, y, g(x))
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(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:
G(s(x)) → G(x)
F(g(x), s(0), y) → F(y, y, g(x))
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → G(x)
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
G(s(x)) → G(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:
POL(G(x1)) = (2)x_1
POL(s(x1)) = 4 + (4)x_1
The value of delta used in the strict ordering is 8.
The following usable rules [17] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(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:
F(g(x), s(0), y) → F(y, y, g(x))
The TRS R consists of the following rules:
f(g(x), s(0), y) → f(y, y, g(x))
g(s(x)) → s(g(x))
g(0) → 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.