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.