Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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 [15] 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 [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

P(s(s(x))) → P(s(x))
FAC(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
          ↳ EdgeDeletionProof

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

P(s(s(x))) → P(s(x))
FAC(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.
We deleted some edges using various graph approximations

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

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

P(s(s(x))) → P(s(x))
FAC(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 [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
QDP
                    ↳ QDPOrderProof
                  ↳ 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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


P(s(s(x))) → P(s(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1)  =  P(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
s1 > P1

Status:
P1: [1]
s1: [1]

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
QDP
                        ↳ PisEmptyProof
                  ↳ QDP

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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.