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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.

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), y) → TIMES(s(x), y)
FAC(s(x), y) → P(s(x))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(p(s(x)), y)
FACTORIAL(x) → FAC(x, s(0))
PLUS(s(x), y) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.
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:

FAC(s(x), y) → TIMES(s(x), y)
FAC(s(x), y) → P(s(x))
TIMES(s(x), y) → P(s(x))
TIMES(s(x), y) → PLUS(y, times(p(s(x)), y))
FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))
P(s(s(x))) → P(s(x))
PLUS(s(x), y) → PLUS(p(s(x)), y)
FACTORIAL(x) → FAC(x, s(0))
PLUS(s(x), y) → P(s(x))
TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP
          ↳ 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:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.
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.


P(s(s(x))) → P(s(x))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(P(x1)) = (2)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

PLUS(s(x), y) → PLUS(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

TIMES(s(x), y) → TIMES(p(s(x)), y)

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
          ↳ QDP
          ↳ QDP

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

FAC(s(x), y) → FAC(p(s(x)), times(s(x), y))

The TRS R consists of the following rules:

plus(0, x) → x
plus(s(x), y) → s(plus(p(s(x)), y))
times(0, y) → 0
times(s(x), y) → plus(y, times(p(s(x)), y))
p(s(0)) → 0
p(s(s(x))) → s(p(s(x)))
fac(0, x) → x
fac(s(x), y) → fac(p(s(x)), times(s(x), y))
factorial(x) → fac(x, s(0))

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