Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/TRCSRSLASHEx4_Zan97

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(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

Q is empty.

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

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
G₁(s₁(X)) -> G₁(X)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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:

SEL₂(s₁(X), cons₂(Y, Z)) -> ACTIVATE₁(Z)
ACTIVATE₁(n__g₁(X)) -> G₁(activate₁(X))
SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))
ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
G₁(s₁(X)) -> G₁(X)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

          ↳ QDP

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

G₁(s₁(X)) -> G₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

G₁(s₁(X)) -> G₁(X)

Used argument filtering: G₁(x1) = x1
s₁(x1) = s₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__g₁(x1) = x1
n__f₁(x1) = n__f₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

          ↳ QDP

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

ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

ACTIVATE₁(n__g₁(X)) -> ACTIVATE₁(X)

Used argument filtering: ACTIVATE₁(x1) = x1
n__g₁(x1) = n__g₁(x1)
Used ordering: Precedence:

trivial

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

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

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

The TRS R consists of the following rules:

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.

SEL₂(s₁(X), cons₂(Y, Z)) -> SEL₂(X, activate₁(Z))

Used argument filtering: SEL₂(x1, x2) = x1
s₁(x1) = s₁(x1)
activate₁(x1) = activate₁(x1)
n__f₁(x1) = n__f
f₁(x1) = f
n__g₁(x1) = n__g₁(x1)
g₁(x1) = g₁(x1)
0 = 0
cons₂(x1, x2) = cons
Used ordering: Precedence:

activate₁ > f > n_{_f}
activate₁ > f > cons
activate₁ > g₁ > s₁
activate₁ > g₁ > n_{_g₁}

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPAfsSolverProof

              ↳ QDP

                ↳ PisEmptyProof

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

f₁(X) -> cons₂(X, n__f₁(n__g₁(X)))
g₁(0) -> s₁(0)
g₁(s₁(X)) -> s₁(s₁(g₁(X)))
sel₂(0, cons₂(X, Y)) -> X
sel₂(s₁(X), cons₂(Y, Z)) -> sel₂(X, activate₁(Z))
f₁(X) -> n__f₁(X)
g₁(X) -> n__g₁(X)
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__g₁(X)) -> g₁(activate₁(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.