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(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(X))
activate(X) → X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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:

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__g(X)) → G(activate(X))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__g(X)) → ACTIVATE(X)
G(s(X)) → G(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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:

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__g(X)) → G(activate(X))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__g(X)) → ACTIVATE(X)
G(s(X)) → G(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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:

SEL(s(X), cons(Y, Z)) → ACTIVATE(Z)
ACTIVATE(n__g(X)) → G(activate(X))
SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
ACTIVATE(n__g(X)) → ACTIVATE(X)
G(s(X)) → G(X)
ACTIVATE(n__f(X)) → F(activate(X))
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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 3 SCCs with 3 less nodes.

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

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

G(s(X)) → G(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.


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

Recursive path order with status [2].
Precedence:
s1 > G1

Status:
G1: multiset
s1: multiset

The following usable rules [14] were oriented: none



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

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.

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

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

ACTIVATE(n__g(X)) → ACTIVATE(X)
ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.


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

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

Recursive path order with status [2].
Precedence:
trivial

Status:
ACTIVATE1: multiset
ng1: multiset

The following usable rules [14] were oriented: none



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

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

ACTIVATE(n__f(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.


ACTIVATE(n__f(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__f(x1)  =  n__f(x1)

Recursive path order with status [2].
Precedence:
nf1 > ACTIVATE1

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
QDP
                        ↳ PisEmptyProof
              ↳ QDP

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.

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

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

SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))

The TRS R consists of the following rules:

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.


SEL(s(X), cons(Y, Z)) → SEL(X, activate(Z))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
SEL(x1, x2)  =  SEL(x1)
s(x1)  =  s(x1)
cons(x1, x2)  =  cons(x1, x2)
activate(x1)  =  activate(x1)
g(x1)  =  x1
0  =  0
n__g(x1)  =  n__g
f(x1)  =  x1
n__f(x1)  =  n__f

Recursive path order with status [2].
Precedence:
cons2 > SEL1 > activate1 > s1
0 > s1
ng > activate1 > s1
nf > activate1 > s1

Status:
SEL1: multiset
nf: multiset
ng: multiset
0: multiset
s1: multiset
cons2: multiset
activate1: multiset

The following usable rules [14] were oriented: none



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

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

f(X) → cons(X, n__f(n__g(X)))
g(0) → s(0)
g(s(X)) → s(s(g(X)))
sel(0, cons(X, Y)) → X
sel(s(X), cons(Y, Z)) → sel(X, activate(Z))
f(X) → n__f(X)
g(X) → n__g(X)
activate(n__f(X)) → f(activate(X))
activate(n__g(X)) → g(activate(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.