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:

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

Q is empty.


↳ QTRS
  ↳ Overlay + Local Confluence

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

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

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

The set Q consists of the following terms:

p(s(x0))
fac(0)
fac(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))
FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

p(s(x0))
fac(0)
fac(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))
FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ 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:

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

The set Q consists of the following terms:

p(s(x0))
fac(0)
fac(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
            ↳ 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(x)) → x

The set Q consists of the following terms:

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



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ 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(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 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(x)) → x

Used ordering: POLO with Polynomial interpretation [25]:

POL(FAC(x1)) = x1   
POL(p(x1)) = x1   
POL(s(x1)) = 1 + x1   



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

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

FAC(s(x)) → FAC(p(s(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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.