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

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

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(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:

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(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:

FAC(s(x)) → P(s(x))
P(s(s(x))) → P(s(x))
FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(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:

FAC(s(x)) → P(s(x))
P(s(s(x))) → P(s(x))
FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

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

P(s(s(x))) → P(s(x))

The TRS R consists of the following rules:

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(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

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

P(s(s(x))) → P(s(x))

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

fac(s(x0))
p(s(0))
p(s(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.

fac(s(x0))
p(s(0))
p(s(s(x0)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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

P(s(s(x))) → P(s(x))

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
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

fac(s(x)) → *(fac(p(s(x))), s(x))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(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
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

fac(s(x0))
p(s(0))
p(s(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.

fac(s(x0))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ RuleRemovalProof

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(0))
p(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented rules of the TRS R:

p(s(0)) → 0

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 1   
POL(FAC(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 2·x1   



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ QDPOrderProof

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

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(0))
p(s(s(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


FAC(s(x)) → FAC(p(s(x)))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
M( p(x1) ) =
/1\
\0/
+
/11\
\10/
·x1

M( s(x1) ) =
/0\
\1/
+
/01\
\01/
·x1

Tuple symbols:
M( FAC(x1) ) = 0+
[0,1]
·x1


Matrix type:
We used a basic matrix type which is not further parametrizeable.


As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:

p(s(s(x))) → s(p(s(x)))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ RuleRemovalProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ PisEmptyProof

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

p(s(s(x))) → s(p(s(x)))

The set Q consists of the following terms:

p(s(0))
p(s(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.