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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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:

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

The TRS R consists of the following rules:

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

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:

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

The TRS R consists of the following rules:

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

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

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

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
            ↳ QDPOrderProof
          ↳ QDP
          ↳ QDP

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

PLUS(x, s(y)) → PLUS(x, y)

The TRS R consists of the following rules:

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

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.


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

POL(PLUS(x1, x2)) = (2)x_2   
POL(s(x1)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

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

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
            ↳ QDPOrderProof
          ↳ QDP

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

TIMES(s(x), y) → TIMES(x, y)

The TRS R consists of the following rules:

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

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.


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

POL(TIMES(x1, x2)) = (2)x_1   
POL(s(x1)) = 1/4 + (7/2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



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

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

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

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:

FAC(s(x)) → FAC(p(s(x)))

The TRS R consists of the following rules:

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

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