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(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
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:
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
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:
F1(s1(x)) -> P1(s1(x))
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
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 1 less node.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(x)) -> F1(p1(s1(x)))
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
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(p1(s1(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:
POL( F1(x1) ) = max{0, 2x1 - 3}
POL( s1(x1) ) = 2x1 + 3
POL( p1(x1) ) = max{0, x1 - 3}
The following usable rules [14] were oriented:
p1(s1(x)) -> x
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(s1(x)) -> s1(s1(f1(p1(s1(x)))))
f1(0) -> 0
p1(s1(x)) -> x
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.