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

The set Q consists of the following terms:

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



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:

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


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:

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

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:

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:

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

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
  ↳ 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:

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

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
  ↳ 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:

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

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.

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



↳ 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:



↳ 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:

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

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
  ↳ 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:

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

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.

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



↳ 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)

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:



↳ 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:

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

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
  ↳ 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:

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

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.

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



↳ 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
  ↳ 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
  ↳ 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
  ↳ 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
  ↳ 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:




Rewriting sequence

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