Termination w.r.t. Q proof of TRS/Ste92perfect.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:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

F(s(x), s(y), z, u) → F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) → F(x, u, z, u)
PERFECTP(s(x)) → F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The TRS R consists of the following rules:

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(x1), x2, x3)

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.

F(s(x), s(y), z, u) → F(x, u, z, u)
F(s(x), 0, z, u) → F(x, u, minus(z, s(x)), u)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
F(x1, x2, x3, x4) = x1
s(x1) = s(x1)
0 = 0
minus(x1, x2) = minus(x1)

Recursive Path Order [2].
Precedence:

0 > s₁ > minus₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ PisEmptyProof

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

perfectp(0) → false
perfectp(s(x)) → f(x, s(0), s(x), s(x))
f(0, y, 0, u) → true
f(0, y, s(z), u) → false
f(s(x), 0, z, u) → f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) → if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

The set Q consists of the following terms:

perfectp(0)
perfectp(s(x0))
f(0, x0, 0, x1)
f(0, x0, s(x1), x2)
f(s(x0), 0, x1, x2)
f(s(x0), s(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.