Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar1/Ste92SLASHperfect.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:

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

  ↳ Non-Overlap Check

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

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.

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:

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))

The set Q consists of the following terms:

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), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

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))

The set Q consists of the following terms:

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), minus₂(y, x), z, u)
F₄(s₁(x), s₁(y), z, u) -> F₄(x, u, z, u)
PERFECTP₁(s₁(x)) -> F₄(x, s₁(0), s₁(x), s₁(x))
F₄(s₁(x), 0, z, u) -> F₄(x, u, minus₂(z, s₁(x)), u)

The TRS R consists of the following rules:

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))

The set Q consists of the following terms:

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 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

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

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:

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))

The set Q consists of the following terms:

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, minus₂(z, s₁(x)), u)

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

trivial

↳ QTRS

  ↳ Non-Overlap Check

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ QDP

              ↳ QDPAfsSolverProof

                ↳ QDP

                  ↳ PisEmptyProof

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

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))

The set Q consists of the following terms:

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.