Termination w.r.t. Q proof of /tmp/tmpLOKzMo/ExIntrod

Termination w.r.t. Q of the following Term Rewriting System could be disproven:

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 [19] 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 [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

FACT(X) → PROD(X, fact(p(X)))
PROD(s(X), Y) → ADD(Y, prod(X, Y))
FACT(X) → P(X)
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → FACT(p(X))
PROD(s(X), Y) → PROD(X, Y)
FACT(X) → ZERO(X)
FACT(X) → IF(zero(X), s(0), prod(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.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

ADD(s(X), Y) → ADD(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

ADD(s(X), Y) → ADD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

PROD(s(X), Y) → PROD(X, Y)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

PROD(s(X), Y) → PROD(X, Y)

From the DPs we obtained the following set of size-change graphs:

PROD(s(X), Y) → PROD(X, Y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

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

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

p(s(X)) → X

The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

p(s(X)) → X

Used ordering: POLO with Polynomial interpretation [25]:

POL(FACT(x₁)) = 2·x₁
POL(p(x₁)) = x₁
POL(s(x₁)) = x₁

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ Instantiation

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

FACT(X) → FACT(p(X))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FACT(X) → FACT(p(X)) we obtained the following new rules:

FACT(p(z0)) → FACT(p(p(z0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

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

FACT(p(z0)) → FACT(p(p(z0)))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule FACT(p(z0)) → FACT(p(p(z0))) we obtained the following new rules:

FACT(p(p(z0))) → FACT(p(p(p(z0))))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ QReductionProof

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

FACT(p(p(z0))) → FACT(p(p(p(z0))))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.

p(s(x0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ UsableRulesReductionPairsProof

                          ↳ QDP

                            ↳ Instantiation

                              ↳ QDP

                                ↳ Instantiation

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ NonTerminationProof

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

FACT(p(p(z0))) → FACT(p(p(p(z0))))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

The TRS P consists of the following rules:

FACT(p(p(z0))) → FACT(p(p(p(z0))))

The TRS R consists of the following rules:none

s = FACT(p(p(z0))) evaluates to t =FACT(p(p(p(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

Semiunifier: [ ]
Matcher: [z0 / p(z0)]

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FACT(p(p(z0))) to FACT(p(p(p(z0)))).