Termination w.r.t. Q proof of TRS/Thiemannfactorial1.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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:

FAC₂(s₁(x), y) -> FAC₂(p₁(s₁(x)), times₂(s₁(x), y))
PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(s₁(x), y) -> P₁(s₁(x))
P₁(s₁(s₁(x))) -> P₁(s₁(x))
TIMES₂(s₁(x), y) -> TIMES₂(p₁(s₁(x)), y)
TIMES₂(s₁(x), y) -> P₁(s₁(x))
FACTORIAL₁(x) -> FAC₂(x, s₁(0))
FAC₂(s₁(x), y) -> P₁(s₁(x))
FAC₂(s₁(x), y) -> TIMES₂(s₁(x), y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(p₁(s₁(x)), y))

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(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:

FAC₂(s₁(x), y) -> FAC₂(p₁(s₁(x)), times₂(s₁(x), y))
PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)
PLUS₂(s₁(x), y) -> P₁(s₁(x))
P₁(s₁(s₁(x))) -> P₁(s₁(x))
TIMES₂(s₁(x), y) -> TIMES₂(p₁(s₁(x)), y)
TIMES₂(s₁(x), y) -> P₁(s₁(x))
FACTORIAL₁(x) -> FAC₂(x, s₁(0))
FAC₂(s₁(x), y) -> P₁(s₁(x))
FAC₂(s₁(x), y) -> TIMES₂(s₁(x), y)
TIMES₂(s₁(x), y) -> PLUS₂(y, times₂(p₁(s₁(x)), y))

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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 6 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(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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 Order [17,21] with Interpretation:

POL( P₁(x₁) ) = max{0, x₁ - 1}

POL( s₁(x₁) ) = x₁ + 1

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(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(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

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

PLUS₂(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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(s₁(x), y) -> PLUS₂(p₁(s₁(x)), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( PLUS₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

The following usable rules [14] were oriented:

p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0

↳ 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(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(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

            ↳ QDPOrderProof

          ↳ QDP

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

TIMES₂(s₁(x), y) -> TIMES₂(p₁(s₁(x)), y)

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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(p₁(s₁(x)), y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( 0 ) = 1

POL( s₁(x₁) ) = x₁ + 1

POL( TIMES₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

The following usable rules [14] were oriented:

p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0

↳ 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(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(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

            ↳ QDPOrderProof

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

FAC₂(s₁(x), y) -> FAC₂(p₁(s₁(x)), times₂(s₁(x), y))

The TRS R consists of the following rules:

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(0))

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.

FAC₂(s₁(x), y) -> FAC₂(p₁(s₁(x)), times₂(s₁(x), y))

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( s₁(x₁) ) = x₁ + 1

POL( 0 ) = 1

POL( FAC₂(x₁, x₂) ) = x₁

POL( times₂(x₁, x₂) ) = max{0, x₂ - 1}

POL( plus₂(x₁, x₂) ) = x₁ + x₂ + 1

POL( p₁(x₁) ) = max{0, x₁ - 1}

The following usable rules [14] were oriented:

p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
p₁(s₁(0)) -> 0

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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

plus₂(0, x) -> x
plus₂(s₁(x), y) -> s₁(plus₂(p₁(s₁(x)), y))
times₂(0, y) -> 0
times₂(s₁(x), y) -> plus₂(y, times₂(p₁(s₁(x)), y))
p₁(s₁(0)) -> 0
p₁(s₁(s₁(x))) -> s₁(p₁(s₁(x)))
fac₂(0, x) -> x
fac₂(s₁(x), y) -> fac₂(p₁(s₁(x)), times₂(s₁(x), y))
factorial₁(x) -> fac₂(x, s₁(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.