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:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
Q is empty.
↳ QTRS
↳ Non-Overlap Check
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
Q is empty.
The TRS is non-overlapping. Hence, we can switch to innermost.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
The set Q consists of the following terms:
g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)
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)) -> F1(x)
F1(s1(x)) -> G1(x)
The TRS R consists of the following rules:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
The set Q consists of the following terms:
g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
G1(s1(x)) -> F1(x)
F1(s1(x)) -> G1(x)
The TRS R consists of the following rules:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
The set Q consists of the following terms:
g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].
The following pairs can be strictly oriented and are deleted.
F1(s1(x)) -> G1(x)
The remaining pairs can at least by weakly be oriented.
G1(s1(x)) -> F1(x)
Used ordering: Combined order from the following AFS and order.
G1(x1) = x1
s1(x1) = s1(x1)
F1(x1) = F1(x1)
Lexicographic Path Order [19].
Precedence:
[s1, F1]
The following usable rules [14] were oriented:
none
↳ QTRS
↳ Non-Overlap Check
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
G1(s1(x)) -> F1(x)
The TRS R consists of the following rules:
g1(s1(x)) -> f1(x)
f1(0) -> s1(0)
f1(s1(x)) -> s1(s1(g1(x)))
g1(0) -> 0
The set Q consists of the following terms:
g1(s1(x0))
f1(0)
f1(s1(x0))
g1(0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.