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:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(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:

ACTIVATE(n__f(X)) → F(X)
F(s(0)) → P(s(0))
F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__f(X)) → F(X)
F(s(0)) → P(s(0))
F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

Q DP problem:
The TRS P consists of the following rules:

ACTIVATE(n__f(X)) → F(X)
F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

Q DP problem:
The TRS P consists of the following rules:

F(s(0)) → F(p(s(0)))

The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(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.


F(s(0)) → F(p(s(0)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
s(x1)  =  s
0  =  0
p(x1)  =  p

Recursive path order with status [2].
Quasi-Precedence:
s > p > 0

Status:
trivial


The following usable rules [14] were oriented:

p(s(0)) → 0



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f(0) → cons(0, n__f(s(0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
activate(n__f(X)) → f(X)
activate(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.