Termination w.r.t. Q proof of TRS/AProVEfac.trs

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:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.

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

TIMES₂(s₁(x), y) -> PLUS₂(times₂(x, y), y)
P₁(s₁(s₁(x))) -> P₁(s₁(x))
FAC₁(s₁(x)) -> P₁(s₁(x))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
FAC₁(s₁(x)) -> TIMES₂(fac₁(p₁(s₁(x))), s₁(x))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(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:

TIMES₂(s₁(x), y) -> PLUS₂(times₂(x, y), y)
P₁(s₁(s₁(x))) -> P₁(s₁(x))
FAC₁(s₁(x)) -> P₁(s₁(x))
TIMES₂(s₁(x), y) -> TIMES₂(x, y)
FAC₁(s₁(x)) -> TIMES₂(fac₁(p₁(s₁(x))), s₁(x))
PLUS₂(x, s₁(y)) -> PLUS₂(x, y)
FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] 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:

P₁(s₁(s₁(x))) -> P₁(s₁(x))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.
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.

P₁(s₁(s₁(x))) -> P₁(s₁(x))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(P₁(x₁)) = x₁²
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] 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:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(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:

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.
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.

PLUS₂(x, s₁(y)) -> PLUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS₂(x₁, x₂)) = x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] 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:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(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:

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

Q is empty.
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.

TIMES₂(s₁(x), y) -> TIMES₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(TIMES₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] 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:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(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:

FAC₁(s₁(x)) -> FAC₁(p₁(s₁(x)))

The TRS R consists of the following rules:

plus₂(x, 0) -> x
plus₂(x, s₁(y)) -> s₁(plus₂(x, y))
times₂(0, y) -> 0
times₂(x, 0) -> 0
times₂(s₁(x), y) -> plus₂(times₂(x, y), y)
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0
fac₁(s₁(x)) -> times₂(fac₁(p₁(s₁(x))), s₁(x))

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