Termination w.r.t. Q proof of TRS/TRCSRExConc_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) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(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:

ACTIVATE₁(n__h₁(X)) -> H₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__h₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(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:

ACTIVATE₁(n__h₁(X)) -> H₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__h₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

ACTIVATE₁(n__h₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(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.

ACTIVATE₁(n__h₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(n__f₁(x₁)) = 3·x₁
POL(n__h₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(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.

ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = 2·x₁
POL(n__f₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

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

f₁(X) -> g₁(n__h₁(n__f₁(X)))
h₁(X) -> n__h₁(X)
f₁(X) -> n__f₁(X)
activate₁(n__h₁(X)) -> h₁(activate₁(X))
activate₁(n__f₁(X)) -> f₁(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.