Termination w.r.t. Q proof of TRS/TRCSRExIntrod_Zan97

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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:

ACTIVATE(n__fact(X)) → ACTIVATE(X)
ADD(s(X), Y) → ADD(X, Y)
PROD(s(X), Y) → ADD(Y, prod(X, Y))
ACTIVATE(n__fact(X)) → FACT(activate(X))
ACTIVATE(n__0) → 0¹
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__prod(X1, X2)) → PROD(activate(X1), activate(X2))
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X2)
FACT(X) → ZERO(X)
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X1)
IF(false, X, Y) → ACTIVATE(Y)
FACT(X) → IF(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
ACTIVATE(n__p(X)) → P(activate(X))
PROD(s(X), Y) → PROD(X, Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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:

ACTIVATE(n__fact(X)) → ACTIVATE(X)
ADD(s(X), Y) → ADD(X, Y)
PROD(s(X), Y) → ADD(Y, prod(X, Y))
ACTIVATE(n__fact(X)) → FACT(activate(X))
ACTIVATE(n__0) → 0¹
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__prod(X1, X2)) → PROD(activate(X1), activate(X2))
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X2)
FACT(X) → ZERO(X)
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X1)
IF(false, X, Y) → ACTIVATE(Y)
FACT(X) → IF(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
ACTIVATE(n__p(X)) → P(activate(X))
PROD(s(X), Y) → PROD(X, Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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:

ACTIVATE(n__fact(X)) → ACTIVATE(X)
PROD(s(X), Y) → ADD(Y, prod(X, Y))
ADD(s(X), Y) → ADD(X, Y)
ACTIVATE(n__0) → 0¹
ACTIVATE(n__fact(X)) → FACT(activate(X))
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__prod(X1, X2)) → PROD(activate(X1), activate(X2))
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X2)
FACT(X) → ZERO(X)
ACTIVATE(n__s(X)) → S(activate(X))
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X1)
IF(false, X, Y) → ACTIVATE(Y)
FACT(X) → IF(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
ACTIVATE(n__p(X)) → P(activate(X))
ACTIVATE(n__s(X)) → ACTIVATE(X)
PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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.

ADD(s(X), Y) → ADD(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
ADD(x1, x2) = x1
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

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:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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:

PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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.

PROD(s(X), Y) → PROD(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
PROD(x1, x2) = x1
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(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

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

ACTIVATE(n__fact(X)) → ACTIVATE(X)
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__prod(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__fact(X)) → FACT(activate(X))
IF(false, X, Y) → ACTIVATE(Y)
FACT(X) → IF(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

fact(X) → if(zero(X), n__s(n__0), n__prod(X, n__fact(n__p(X))))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
prod(0, X) → 0
prod(s(X), Y) → add(Y, prod(X, Y))
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
prod(X1, X2) → n__prod(X1, X2)
fact(X) → n__fact(X)
p(X) → n__p(X)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__prod(X1, X2)) → prod(activate(X1), activate(X2))
activate(n__fact(X)) → fact(activate(X))
activate(n__p(X)) → p(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.