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:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(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:

P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(x0)))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

P1(s1(s1(x))) -> P1(s1(x))
FAC1(s1(x)) -> P1(s1(x))
FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(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
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

P1(s1(s1(x))) -> P1(s1(x))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(x0)))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


P1(s1(s1(x))) -> P1(s1(x))
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
P1(x1)  =  x1
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
trivial

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

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

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(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
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

FAC1(s1(x)) -> FAC1(p1(s1(x)))

The TRS R consists of the following rules:

fac1(s1(x)) -> *2(fac1(p1(s1(x))), s1(x))
p1(s1(0)) -> 0
p1(s1(s1(x))) -> s1(p1(s1(x)))

The set Q consists of the following terms:

fac1(s1(x0))
p1(s1(0))
p1(s1(s1(x0)))

We have to consider all minimal (P,Q,R)-chains.