Termination w.r.t. Q proof of TRS/secret06divExp.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

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:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

DIV(s(x), s(y)) → MINUS(x, y)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
F(x, s(y), b) → MINUS(s(y), s(0))
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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:

DIV(s(x), s(y)) → MINUS(x, y)
DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
F(x, s(y), b) → MINUS(s(y), s(0))
MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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:

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))
DIV(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → DIV(f(x, minus(s(y), s(0)), b), b)
F(x, s(y), b) → F(x, minus(s(y), s(0)), b)
MINUS(s(x), s(y)) → MINUS(x, y)
F(x, s(y), b) → MINUS(s(y), s(0))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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.

MINUS(s(x), s(y)) → MINUS(x, y)

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

Recursive Path Order [2].
Precedence:

s₁ > MINUS₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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.

DIV(s(x), s(y)) → DIV(minus(x, y), s(y))

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

Recursive Path Order [2].
Precedence:

[DIV₂, s_1] > 0

The following usable rules [14] were oriented:

minus(x, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

F(x, s(y), b) → F(x, minus(s(y), s(0)), b)

The TRS R consists of the following rules:

minus(x, x) → 0
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
div(s(x), s(y)) → s(div(minus(x, y), s(y)))
div(0, s(y)) → 0
f(x, 0, b) → x
f(x, s(y), b) → div(f(x, minus(s(y), s(0)), b), b)

The set Q consists of the following terms:

minus(x0, x0)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
div(s(x0), s(x1))
div(0, s(x0))
f(x0, 0, x1)
f(x0, s(x1), x2)

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