Termination w.r.t. Q proof of /tmp/tmpjHvYqd/div

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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

PLUS(0, s(x)) → PLUS(0, x)
IFY(true, x, y) → IF(ge(x, y), x, y)
DIV(x, y) → GE(y, s(0))
MINUS(s(x), s(y)) → MINUS(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
IF(true, x, y) → MINUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
GE(s(x), s(y)) → GE(x, y)
GE(s(x), 0) → GE(x, 0)
GE(0, s(s(x))) → GE(0, s(x))
PLUS(s(x), y) → PLUS(x, y)
IFY(true, x, y) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

PLUS(0, s(x)) → PLUS(0, x)
IFY(true, x, y) → IF(ge(x, y), x, y)
DIV(x, y) → GE(y, s(0))
MINUS(s(x), s(y)) → MINUS(x, y)
IF(true, x, y) → DIV(minus(x, y), y)
MINUS(0, s(x)) → MINUS(0, x)
MINUS(s(x), 0) → MINUS(x, 0)
IF(true, x, y) → MINUS(x, y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
GE(s(x), s(y)) → GE(x, y)
GE(s(x), 0) → GE(x, 0)
GE(0, s(s(x))) → GE(0, s(x))
PLUS(s(x), y) → PLUS(x, y)
IFY(true, x, y) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 9 SCCs with 3 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(0, s(x)) → PLUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(0, s(x)) → PLUS(0, x)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(0, s(x)) → PLUS(0, x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

PLUS(0, s(x)) → PLUS(0, x)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, y)

From the DPs we obtained the following set of size-change graphs:

PLUS(s(x), y) → PLUS(x, y)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), 0) → MINUS(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), 0) → MINUS(x, 0)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), 0) → MINUS(x, 0)

From the DPs we obtained the following set of size-change graphs:

MINUS(s(x), 0) → MINUS(x, 0)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(0, s(x)) → MINUS(0, x)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(0, s(x)) → MINUS(0, x)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(0, s(x)) → MINUS(0, x)

From the DPs we obtained the following set of size-change graphs:

MINUS(0, s(x)) → MINUS(0, x)
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ 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:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

From the DPs we obtained the following set of size-change graphs:

MINUS(s(x), s(y)) → MINUS(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

GE(0, s(s(x))) → GE(0, s(x))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

GE(0, s(s(x))) → GE(0, s(x))

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

GE(0, s(s(x))) → GE(0, s(x))

From the DPs we obtained the following set of size-change graphs:

GE(0, s(s(x))) → GE(0, s(x))
The graph contains the following edges 1 >= 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

GE(s(x), 0) → GE(x, 0)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

GE(s(x), 0) → GE(x, 0)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

GE(s(x), 0) → GE(x, 0)

From the DPs we obtained the following set of size-change graphs:

GE(s(x), 0) → GE(x, 0)
The graph contains the following edges 1 > 1, 2 >= 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

R is empty.
The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

GE(s(x), s(y)) → GE(x, y)

From the DPs we obtained the following set of size-change graphs:

GE(s(x), s(y)) → GE(x, y)
The graph contains the following edges 1 > 1, 2 > 2

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(0)) → false
ge(0, s(s(x))) → ge(0, s(x))
ge(s(x), s(y)) → ge(x, y)
minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
plus(0, 0) → 0
plus(0, s(x)) → s(plus(0, x))
plus(s(x), y) → s(plus(x, y))
div(x, y) → ify(ge(y, s(0)), x, y)
ify(false, x, y) → divByZeroError
ify(true, x, y) → if(ge(x, y), x, y)
if(false, x, y) → 0
if(true, x, y) → s(div(minus(x, y), y))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))
plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

plus(0, 0)
plus(0, s(x0))
plus(s(x0), x1)
div(x0, x1)
ify(false, x0, x1)
ify(true, x0, x1)
if(false, x0, x1)
if(true, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

IF(true, x, y) → DIV(minus(x, y), y)
DIV(x, y) → IFY(ge(y, s(0)), x, y)
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule DIV(x, y) → IFY(ge(y, s(0)), x, y) at position [0] we obtained the following new rules:

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
DIV(y0, 0) → IFY(false, y0, 0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
DIV(y0, 0) → IFY(false, y0, 0)
IF(true, x, y) → DIV(minus(x, y), y)
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

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

DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IF(true, x, y) → DIV(minus(x, y), y)
IFY(true, x, y) → IF(ge(x, y), x, y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IFY(true, x, y) → IF(ge(x, y), x, y) at position [0] we obtained the following new rules:

IFY(true, 0, 0) → IF(true, 0, 0)
IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))
IFY(true, 0, s(0)) → IF(false, 0, s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IFY(true, 0, 0) → IF(true, 0, 0)
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IF(true, x, y) → DIV(minus(x, y), y)
IFY(true, 0, s(0)) → IF(false, 0, s(0))
IFY(true, s(x0), 0) → IF(ge(x0, 0), s(x0), 0)
IFY(true, 0, s(s(x0))) → IF(ge(0, s(x0)), 0, s(s(x0)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IF(true, x, y) → DIV(minus(x, y), y)

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(true, x, y) → DIV(minus(x, y), y) we obtained the following new rules:

IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF(true, s(z0), s(z1)) → DIV(minus(s(z0), s(z1)), s(z1)) at position [0] we obtained the following new rules:

IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule DIV(y0, s(x0)) → IFY(ge(x0, 0), y0, s(x0)) we obtained the following new rules:

DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF(true, s(z0), s(z1)) → DIV(minus(z0, z1), s(z1)) at position [0] we obtained the following new rules:

IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(0), s(0)) → DIV(0, s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IF(true, s(0), s(s(x0))) → DIV(minus(0, x0), s(s(x0)))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IF(true, s(0), s(0)) → DIV(0, s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule DIV(s(y_1), s(x1)) → IFY(ge(x1, 0), s(y_1), s(x1)) we obtained the following new rules:

DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule DIV(s(y_0), s(0)) → IFY(ge(0, 0), s(y_0), s(0)) at position [0] we obtained the following new rules:

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule DIV(s(x0), s(s(z1))) → IFY(ge(s(z1), 0), s(x0), s(s(z1))) at position [0] we obtained the following new rules:

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

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

IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IFY(true, s(x0), s(x1)) → IF(ge(x0, x1), s(x0), s(x1)) we obtained the following new rules:

IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

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

IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                ↳ QDP

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ ForwardInstantiation

                                                                                ↳ QDP

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))
IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IFY(true, s(z0), s(0)) → IF(ge(z0, 0), s(z0), s(0)) we obtained the following new rules:

IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ ForwardInstantiation

                                                                                        ↳ QDP

                                                                                          ↳ Rewriting

                                                                                ↳ QDP

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

IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0))
DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IFY(true, s(s(y_1)), s(0)) → IF(ge(s(y_1), 0), s(s(y_1)), s(0)) at position [0] we obtained the following new rules:

IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ ForwardInstantiation

                                                                                        ↳ QDP

                                                                                          ↳ Rewriting

                                                                                            ↳ QDP

                                                                                              ↳ RuleRemovalProof

                                                                                ↳ QDP

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

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))
IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
minus(0, 0) → 0
minus(s(x), 0) → s(minus(x, 0))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

DIV(s(y_0), s(0)) → IFY(true, s(y_0), s(0))
IFY(true, s(s(y_1)), s(0)) → IF(ge(y_1, 0), s(s(y_1)), s(0))

Strictly oriented rules of the TRS R:

ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)

Used ordering: POLO with Polynomial interpretation [25]:

POL(0) = 0
POL(DIV(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(IF(x₁, x₂, x₃)) = 2·x₁ + x₂ + x₃
POL(IFY(x₁, x₂, x₃)) = x₁ + 2·x₂ + 2·x₃
POL(ge(x₁, x₂)) = 2 + x₁ + x₂
POL(minus(x₁, x₂)) = x₁ + 2·x₂
POL(s(x₁)) = 1 + 2·x₁
POL(true) = 1

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                  ↳ UsableRulesProof

                                                                                    ↳ QDP

                                                                                      ↳ ForwardInstantiation

                                                                                        ↳ QDP

                                                                                          ↳ Rewriting

                                                                                            ↳ QDP

                                                                                              ↳ RuleRemovalProof

                                                                                                ↳ QDP

                                                                                                  ↳ DependencyGraphProof

                                                                                ↳ QDP

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

IF(true, s(s(x0)), s(0)) → DIV(s(minus(x0, 0)), s(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

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

IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1)))
DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IFY(true, s(z0), s(s(z1))) → IF(ge(z0, s(z1)), s(z0), s(s(z1))) we obtained the following new rules:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(s(y_1), s(x1)), s(s(y_1)), s(s(x1))) at position [0] we obtained the following new rules:

IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ QDPOrderProof

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))
IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IF(true, s(s(x0)), s(s(x1))) → DIV(minus(x0, x1), s(s(x1)))

The remaining pairs can at least be oriented weakly.

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(DIV(x₁, x₂)) = x₁ + x₂
POL(IF(x₁, x₂, x₃)) = x₂ + x₃
POL(IFY(x₁, x₂, x₃)) = x₂ + x₃
POL(false) = 0
POL(ge(x₁, x₂)) = 0
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁
POL(true) = 0

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Narrowing

                                  ↳ QDP

                                    ↳ DependencyGraphProof

                                      ↳ QDP

                                        ↳ Instantiation

                                          ↳ QDP

                                            ↳ Rewriting

                                              ↳ QDP

                                                ↳ ForwardInstantiation

                                                  ↳ QDP

                                                    ↳ Narrowing

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ Instantiation

                                                              ↳ QDP

                                                                ↳ Rewriting

                                                                  ↳ QDP

                                                                    ↳ Rewriting

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ DependencyGraphProof

                                                                              ↳ AND

                                                                                ↳ QDP

                                                                                ↳ QDP

                                                                                  ↳ ForwardInstantiation

                                                                                    ↳ QDP

                                                                                      ↳ Rewriting

                                                                                        ↳ QDP

                                                                                          ↳ QDPOrderProof

                                                                                            ↳ QDP

                                                                                              ↳ DependencyGraphProof

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

DIV(s(x0), s(s(z1))) → IFY(ge(z1, 0), s(x0), s(s(z1)))
IFY(true, s(s(y_1)), s(s(x1))) → IF(ge(y_1, x1), s(s(y_1)), s(s(x1)))

The TRS R consists of the following rules:

minus(0, 0) → 0
minus(0, s(x)) → minus(0, x)
minus(s(x), 0) → s(minus(x, 0))
minus(s(x), s(y)) → minus(x, y)
ge(0, s(0)) → false
ge(s(x), s(y)) → ge(x, y)
ge(0, 0) → true
ge(s(x), 0) → ge(x, 0)
ge(0, s(s(x))) → ge(0, s(x))

The set Q consists of the following terms:

ge(0, 0)
ge(s(x0), 0)
ge(0, s(0))
ge(0, s(s(x0)))
ge(s(x0), s(x1))
minus(0, 0)
minus(0, s(x0))
minus(s(x0), 0)
minus(s(x0), s(x1))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 2 less nodes.