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:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
Q is empty.
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
Q is empty.
The TRS is overlay and locally confluent. By [15] we can switch to innermost.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
The set Q consists of the following terms:
g(s(x0))
f(0)
f(s(x0))
g(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:
G(s(x)) → F(x)
F(s(x)) → G(x)
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
The set Q consists of the following terms:
g(s(x0))
f(0)
f(s(x0))
g(0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → F(x)
F(s(x)) → G(x)
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
The set Q consists of the following terms:
g(s(x0))
f(0)
f(s(x0))
g(0)
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
G(s(x)) → F(x)
F(s(x)) → G(x)
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
The set Q consists of the following terms:
g(s(x0))
f(0)
f(s(x0))
g(0)
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.
G(s(x)) → F(x)
F(s(x)) → G(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
G(x1) = x1
s(x1) = s(x1)
F(x1) = F(x1)
Lexicographic Path Order [19].
Precedence:
s1 > F1
The following usable rules [14] were oriented:
none
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ EdgeDeletionProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
g(s(x)) → f(x)
f(0) → s(0)
f(s(x)) → s(s(g(x)))
g(0) → 0
The set Q consists of the following terms:
g(s(x0))
f(0)
f(s(x0))
g(0)
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.