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(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(0)))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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(0)) -> F1(s1(0))
G1(s1(0)) -> G1(f1(s1(0)))
F1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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(0)) -> F1(s1(0))
G1(s1(0)) -> G1(f1(s1(0)))
F1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(0)))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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.
F1(s1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:
POL(F1(x1)) = 3·x1
POL(s1(x1)) = 1 + x1
The following usable rules [14] were oriented:
none
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(f1(x)) -> f1(x)
f1(s1(x)) -> f1(x)
g1(s1(0)) -> g1(f1(s1(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.