Termination w.r.t. Q proof of TRS/nontermin/CSRExIntrod

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), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FACT(X) → ZERO(X)
FACT(X) → PROD(X, fact(p(X)))
PROD(s(X), Y) → ADD(Y, prod(X, Y))
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → P(X)
FACT(X) → IF(zero(X), s(0), prod(X, fact(p(X))))
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

FACT(X) → ZERO(X)
FACT(X) → PROD(X, fact(p(X)))
PROD(s(X), Y) → ADD(Y, prod(X, Y))
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → P(X)
FACT(X) → IF(zero(X), s(0), prod(X, fact(p(X))))
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

FACT(X) → ZERO(X)
FACT(X) → PROD(X, fact(p(X)))
ADD(s(X), Y) → ADD(X, Y)
PROD(s(X), Y) → ADD(Y, prod(X, Y))
FACT(X) → P(X)
FACT(X) → IF(zero(X), s(0), prod(X, fact(p(X))))
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

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) = ADD(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > ADD₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

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) = PROD(x1)
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

s₁ > PROD₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

fact(X) → if(zero(X), s(0), prod(X, fact(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) → X
if(false, X, Y) → Y
zero(0) → true
zero(s(X)) → false
p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

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