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:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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__fact₁(X)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
PROD₂(s₁(X), Y) -> ADD₂(Y, prod₂(X, Y))
ACTIVATE₁(n__fact₁(X)) -> FACT₁(activate₁(X))
ACTIVATE₁(n__0) -> 0¹
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(add₂(X, Y))
ACTIVATE₁(n__prod₂(X1, X2)) -> PROD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X2)
FACT₁(X) -> ZERO₁(X)
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X1)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
FACT₁(X) -> IF₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
ACTIVATE₁(n__p₁(X)) -> P₁(activate₁(X))
PROD₂(s₁(X), Y) -> PROD₂(X, Y)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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__fact₁(X)) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> ADD₂(X, Y)
PROD₂(s₁(X), Y) -> ADD₂(Y, prod₂(X, Y))
ACTIVATE₁(n__fact₁(X)) -> FACT₁(activate₁(X))
ACTIVATE₁(n__0) -> 0¹
IF₃(true, X, Y) -> ACTIVATE₁(X)
ADD₂(s₁(X), Y) -> S₁(add₂(X, Y))
ACTIVATE₁(n__prod₂(X1, X2)) -> PROD₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X2)
FACT₁(X) -> ZERO₁(X)
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X1)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
FACT₁(X) -> IF₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
ACTIVATE₁(n__p₁(X)) -> P₁(activate₁(X))
PROD₂(s₁(X), Y) -> PROD₂(X, Y)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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 3 SCCs with 7 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

The TRS R consists of the following rules:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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.

ADD₂(s₁(X), Y) -> ADD₂(X, Y)

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

POL(ADD₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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

            ↳ QDPOrderProof

          ↳ QDP

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

PROD₂(s₁(X), Y) -> PROD₂(X, Y)

The TRS R consists of the following rules:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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.

PROD₂(s₁(X), Y) -> PROD₂(X, Y)

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

POL(PROD₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(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

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

ACTIVATE₁(n__fact₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X2)
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__prod₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__fact₁(X)) -> FACT₁(activate₁(X))
IF₃(false, X, Y) -> ACTIVATE₁(Y)
FACT₁(X) -> IF₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

fact₁(X) -> if₃(zero₁(X), n__s₁(n__0), n__prod₂(X, n__fact₁(n__p₁(X))))
add₂(0, X) -> X
add₂(s₁(X), Y) -> s₁(add₂(X, Y))
prod₂(0, X) -> 0
prod₂(s₁(X), Y) -> add₂(Y, prod₂(X, Y))
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
zero₁(0) -> true
zero₁(s₁(X)) -> false
p₁(s₁(X)) -> X
s₁(X) -> n__s₁(X)
0 -> n__0
prod₂(X1, X2) -> n__prod₂(X1, X2)
fact₁(X) -> n__fact₁(X)
p₁(X) -> n__p₁(X)
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__0) -> 0
activate₁(n__prod₂(X1, X2)) -> prod₂(activate₁(X1), activate₁(X2))
activate₁(n__fact₁(X)) -> fact₁(activate₁(X))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> X

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