Termination w.r.t. Q proof of TRS/Ste92perfect2.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:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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:

F₄(s₁(x), 0, z, u) -> MINUS₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> MINUS₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), s₁(y), z, u) -> LE₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> IF₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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:

F₄(s₁(x), 0, z, u) -> MINUS₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> MINUS₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), s₁(y), z, u) -> LE₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> IF₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

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

POL( LE₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

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

POL( MINUS₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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

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

F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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.

F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The remaining pairs can at least be oriented weakly.

F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( F₄(x₁, ..., x₄) ) = 3x₁ + 3x₄ + 2

POL( minus₂(x₁, x₂) ) = max{0, 3x₂ - 1}

POL( 0 ) = max{0, -3}

POL( s₁(x₁) ) = 3x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)

The TRS R consists of the following rules:

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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.

F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), minus₂(y, x), z, u)

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

POL( F₄(x₁, ..., x₄) ) = 3x₂ + 2x₄ + 2

POL( minus₂(x₁, x₂) ) = x₁

POL( 0 ) = 2

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented:

minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

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

minus₂(0, y) -> 0
minus₂(s₁(x), 0) -> s₁(x)
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
if₃(true, x, y) -> x
if₃(false, x, y) -> y
perfectp₁(0) -> false
perfectp₁(s₁(x)) -> f₄(x, s₁(0), s₁(x), s₁(x))
f₄(0, y, 0, u) -> true
f₄(0, y, s₁(z), u) -> false
f₄(s₁(x), 0, z, u) -> f₄(x, u, minus₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(le₂(x, y), f₄(s₁(x), minus₂(y, x), z, u), f₄(x, u, z, u))

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.