Termination w.r.t. Q proof of TRS/TRCSREx5_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) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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:

IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)
F₁(X) -> IF₃(X, c, n__f₁(n__true))
ACTIVATE₁(n__true) -> TRUE

The TRS R consists of the following rules:

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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:

IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)
F₁(X) -> IF₃(X, c, n__f₁(n__true))
ACTIVATE₁(n__true) -> TRUE

The TRS R consists of the following rules:

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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 1 less node.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
ACTIVATE₁(n__f₁(X)) -> ACTIVATE₁(X)
F₁(X) -> IF₃(X, c, n__f₁(n__true))

The TRS R consists of the following rules:

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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.

IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
F₁(X) -> IF₃(X, c, n__f₁(n__true))

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = x₁
POL(F₁(x₁)) = 1
POL(IF₃(x₁, x₂, x₃)) = x₃
POL(activate₁(x₁)) = 0
POL(c) = 0
POL(f₁(x₁)) = 0
POL(false) = 0
POL(if₃(x₁, x₂, x₃)) = 0
POL(n__f₁(x₁)) = 1 + x₁
POL(n__true) = 0
POL(true) = 0

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:

IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__f₁(X)) -> F₁(activate₁(X))
F₁(X) -> IF₃(X, c, n__f₁(n__true))

The TRS R consists of the following rules:

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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)) -> F₁(activate₁(X))

The remaining pairs can at least be oriented weakly.

IF₃(false, X, Y) -> ACTIVATE₁(Y)
F₁(X) -> IF₃(X, c, n__f₁(n__true))

Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE₁(x₁)) = 1 + x₁
POL(F₁(x₁)) = x₁
POL(IF₃(x₁, x₂, x₃)) = x₁ + x₃
POL(activate₁(x₁)) = x₁
POL(c) = 0
POL(f₁(x₁)) = x₁
POL(false) = 1
POL(if₃(x₁, x₂, x₃)) = x₁ + x₂ + x₃
POL(n__f₁(x₁)) = x₁
POL(n__true) = 0
POL(true) = 0

The following usable rules [14] were oriented:

if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> if₃(X, c, n__f₁(n__true))
activate₁(n__f₁(X)) -> f₁(activate₁(X))
true -> n__true
f₁(X) -> n__f₁(X)
activate₁(n__true) -> true
if₃(true, X, Y) -> X
activate₁(X) -> X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ DependencyGraphProof

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

IF₃(false, X, Y) -> ACTIVATE₁(Y)
F₁(X) -> IF₃(X, c, n__f₁(n__true))

The TRS R consists of the following rules:

f₁(X) -> if₃(X, c, n__f₁(n__true))
if₃(true, X, Y) -> X
if₃(false, X, Y) -> activate₁(Y)
f₁(X) -> n__f₁(X)
true -> n__true
activate₁(n__f₁(X)) -> f₁(activate₁(X))
activate₁(n__true) -> true
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 0 SCCs with 2 less nodes.