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(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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__s(X)) → S(activate(X))
ACTIVATE(n__0) → 01
ACTIVATE(n__f(X)) → F(activate(X))
F(s(0)) → P(s(0))
F(s(0)) → F(p(s(0)))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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__s(X)) → S(activate(X))
ACTIVATE(n__0) → 01
ACTIVATE(n__f(X)) → F(activate(X))
F(s(0)) → P(s(0))
F(s(0)) → F(p(s(0)))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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__s(X)) → S(activate(X))
ACTIVATE(n__0) → 01
ACTIVATE(n__f(X)) → F(activate(X))
F(s(0)) → F(p(s(0)))
F(s(0)) → P(s(0))
ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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 2 SCCs with 4 less nodes.

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

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(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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)  =  F(x1)
s(x1)  =  s(x1)
0  =  0
p(x1)  =  p
n__s(x1)  =  n__s
n__0  =  n__0

Recursive path order with status [2].
Quasi-Precedence:
[0, p, n0] > F1 > [s1, ns]

Status:
n0: multiset
s1: multiset
0: multiset
ns: multiset
p: []
F1: multiset


The following usable rules [14] were oriented:

p(s(0)) → 0
0n__0



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

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

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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.

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

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

ACTIVATE(n__f(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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.


ACTIVATE(n__f(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.

ACTIVATE(n__s(X)) → ACTIVATE(X)
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__f(x1)  =  n__f(x1)
n__s(x1)  =  x1

Recursive path order with status [2].
Quasi-Precedence:
[ACTIVATE1, nf1]

Status:
ACTIVATE1: multiset
nf1: multiset


The following usable rules [14] were oriented: none



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

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

ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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.


ACTIVATE(n__s(X)) → ACTIVATE(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ACTIVATE(x1)  =  ACTIVATE(x1)
n__s(x1)  =  n__s(x1)

Recursive path order with status [2].
Quasi-Precedence:
[ACTIVATE1, ns1]

Status:
ACTIVATE1: multiset
ns1: multiset


The following usable rules [14] were oriented: none



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

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

f(0) → cons(0, n__f(n__s(n__0)))
f(s(0)) → f(p(s(0)))
p(s(0)) → 0
f(X) → n__f(X)
s(X) → n__s(X)
0n__0
activate(n__f(X)) → f(activate(X))
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
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.