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:
f2(g1(x), s1(0)) -> f2(g1(x), 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:
f2(g1(x), s1(0)) -> f2(g1(x), g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
G1(s1(x)) -> G1(x)
F2(g1(x), s1(0)) -> F2(g1(x), g1(x))
The TRS R consists of the following rules:
f2(g1(x), s1(0)) -> f2(g1(x), 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:
G1(s1(x)) -> G1(x)
F2(g1(x), s1(0)) -> F2(g1(x), g1(x))
The TRS R consists of the following rules:
f2(g1(x), s1(0)) -> f2(g1(x), 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 [13,14,18] contains 2 SCCs.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPOrderProof
↳ 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:
f2(g1(x), s1(0)) -> f2(g1(x), g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be oriented strictly and are deleted.
G1(s1(x)) -> G1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:
POL(G1(x1)) = x1
POL(s1(x1)) = 1 + x1
The following usable rules [14] 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:
f2(g1(x), s1(0)) -> f2(g1(x), 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:
F2(g1(x), s1(0)) -> F2(g1(x), g1(x))
The TRS R consists of the following rules:
f2(g1(x), s1(0)) -> f2(g1(x), g1(x))
g1(s1(x)) -> s1(g1(x))
g1(0) -> 0
Q is empty.
We have to consider all minimal (P,Q,R)-chains.