Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar3/HMSLASHt003.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

<=¹₂(s₁(x), s₁(y)) -> <=¹₂(x, y)

Used argument filtering: <=¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

              ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-¹₂(s₁(x), s₁(y)) -> -¹₂(x, y)

Used argument filtering: -¹₂(x1, x2) = x2
s₁(x1) = s₁(x1)
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)

Used argument filtering: F₄(x1, x2, x3, x4) = x1
s₁(x1) = s₁(x1)
-₂(x1, x2) = x1
0 = 0
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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

Used argument filtering: F₄(x1, x2, x3, x4) = x2
s₁(x1) = s₁(x1)
-₂(x1, x2) = x1
Used ordering: Quasi Precedence: trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ QDPAfsSolverProof

                  ↳ QDP

                    ↳ QDPAfsSolverProof

                      ↳ 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))

The set Q consists of the following terms:

-₂(x0, 0)
-₂(s₁(x0), s₁(x1))
<=₂(0, x0)
<=₂(s₁(x0), 0)
<=₂(s₁(x0), s₁(x1))
if₃(true, x0, x1)
if₃(false, x0, x1)
perfectp₁(0)
perfectp₁(s₁(x0))
f₄(0, x0, 0, x1)
f₄(0, x0, s₁(x1), x2)
f₄(s₁(x0), 0, x1, x2)
f₄(s₁(x0), s₁(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.