Termination w.r.t. Q proof of /tmp/tmp9v_ReK/ExIntrod_Zan97

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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
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(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X

Q is empty.

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

PROD(s(X), Y) → ADD(Y, prod(X, Y))
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__prod(X1, X2)) → PROD(X1, X2)
FACT(X) → IF(zero(X), n__s(0), n__prod(X, fact(p(X))))
ACTIVATE(n__s(X)) → S(X)
FACT(X) → P(X)
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → FACT(p(X))
IF(false, X, Y) → ACTIVATE(Y)
PROD(s(X), Y) → PROD(X, Y)
FACT(X) → ZERO(X)

The TRS R consists of the following rules:

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

PROD(s(X), Y) → ADD(Y, prod(X, Y))
ADD(s(X), Y) → S(add(X, Y))
ACTIVATE(n__prod(X1, X2)) → PROD(X1, X2)
FACT(X) → IF(zero(X), n__s(0), n__prod(X, fact(p(X))))
ACTIVATE(n__s(X)) → S(X)
FACT(X) → P(X)
IF(true, X, Y) → ACTIVATE(X)
ADD(s(X), Y) → ADD(X, Y)
FACT(X) → FACT(p(X))
IF(false, X, Y) → ACTIVATE(Y)
PROD(s(X), Y) → PROD(X, Y)
FACT(X) → ZERO(X)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 9 less nodes.

↳ 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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

  ↳ 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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

  ↳ 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [15] to enlarge Q to all left-hand sides of R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ QDP

                ↳ MNOCProof

                  ↳ 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)))).