Termination w.r.t. Q proof of TRS/HMt003.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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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), s₁(y), z, u) -> F₄(s₁(x), -₂(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, -₂(z, s₁(x)), u)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
<=¹₂(s₁(x), s₁(y)) -> <=¹₂(x, y)
F₄(s₁(x), 0, z, u) -> -¹₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> <=¹₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> -¹₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> IF₃(<=₂(x, y), f₄(s₁(x), -₂(y, x), z, u), f₄(x, u, z, u))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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), s₁(y), z, u) -> F₄(s₁(x), -₂(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, -₂(z, s₁(x)), u)
-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
<=¹₂(s₁(x), s₁(y)) -> <=¹₂(x, y)
F₄(s₁(x), 0, z, u) -> -¹₂(z, s₁(x))
F₄(s₁(x), s₁(y), z, u) -> <=¹₂(x, y)
F₄(s₁(x), s₁(y), z, u) -> -¹₂(y, x)
F₄(s₁(x), s₁(y), z, u) -> IF₃(<=₂(x, y), f₄(s₁(x), -₂(y, x), z, u), f₄(x, u, z, u))

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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:

<=¹₂(s₁(x), s₁(y)) -> <=¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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.

<=¹₂(s₁(x), s₁(y)) -> <=¹₂(x, y)

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

POL(<=¹₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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:

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

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

POL(-¹₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 1 + x₁

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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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), -₂(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, -₂(z, s₁(x)), u)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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, -₂(z, s₁(x)), u)

The remaining pairs can at least be oriented weakly.

F₄(s₁(x), s₁(y), z, u) -> F₄(s₁(x), -₂(y, x), z, u)

Used ordering: Polynomial interpretation [21]:

POL(-₂(x₁, x₂)) = 0
POL(0) = 0
POL(F₄(x₁, x₂, x₃, x₄)) = x₁
POL(s₁(x₁)) = 1 + x₁

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), -₂(y, x), z, u)

The TRS R consists of the following rules:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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), -₂(y, x), z, u)

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

POL(-₂(x₁, x₂)) = x₁
POL(0) = 0
POL(F₄(x₁, x₂, x₃, x₄)) = 1 + x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented:

-₂(s₁(x), s₁(y)) -> -₂(x, y)
-₂(x, 0) -> 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:

-₂(x, 0) -> x
-₂(s₁(x), s₁(y)) -> -₂(x, y)
<=₂(0, y) -> true
<=₂(s₁(x), 0) -> false
<=₂(s₁(x), s₁(y)) -> <=₂(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, -₂(z, s₁(x)), u)
f₄(s₁(x), s₁(y), z, u) -> if₃(<=₂(x, y), f₄(s₁(x), -₂(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.