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(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
f1(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
ACTIVATE1(n__f1(X)) -> F1(X)
F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
The TRS R consists of the following rules:
f1(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(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:
ACTIVATE1(n__f1(X)) -> F1(X)
F1(s1(0)) -> P1(s1(0))
F1(s1(0)) -> F1(p1(s1(0)))
The TRS R consists of the following rules:
f1(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
Q DP problem:
The TRS P consists of the following rules:
F1(s1(0)) -> F1(p1(s1(0)))
The TRS R consists of the following rules:
f1(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
F1(s1(0)) -> F1(p1(s1(0)))
Used argument filtering: F1(x1) = x1
s1(x1) = s
0 = 0
p1(x1) = p
Used ordering: Quasi Precedence:
s > [p, 0]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
Q DP problem:
P is empty.
The TRS R consists of the following rules:
f1(0) -> cons2(0, n__f1(s1(0)))
f1(s1(0)) -> f1(p1(s1(0)))
p1(s1(0)) -> 0
f1(X) -> n__f1(X)
activate1(n__f1(X)) -> f1(X)
activate1(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.