Termination w.r.t. Q proof of /tmp/tmpd0FR

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:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

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:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(true, x, y, z) → P(z)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y, z) → PLUS(z, s(0))
QUOT(x, y, z) → ZERO(x)
PLUS(s(x), y) → PLUS(x, s(y))
IF(false, x, s(y), z) → MINUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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:

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
IF(true, x, y, z) → P(z)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(x, y, z) → PLUS(z, s(0))
QUOT(x, y, z) → ZERO(x)
PLUS(s(x), y) → PLUS(x, s(y))
IF(false, x, s(y), z) → MINUS(x, s(y))
DIV(x, y) → QUOT(x, y, 0)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 3 SCCs with 5 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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

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

PLUS(s(x), y) → PLUS(x, s(y))

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

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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.

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

PLUS(s(x), y) → PLUS(x, s(y))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ 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:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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.

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ 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

                ↳ UsableRulesProof

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
zero(s(x)) → false
zero(0) → true
p(s(x)) → x
div(x, y) → quot(x, y, 0)
quot(x, y, z) → if(zero(x), x, y, plus(z, s(0)))
if(true, x, y, z) → p(z)
if(false, x, s(y), z) → quot(minus(x, s(y)), s(y), z)

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)
p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

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.

p(s(x0))
div(x0, x1)
quot(x0, x1, x2)
if(true, x0, x1, x2)
if(false, x0, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(x, y, z) → IF(zero(x), x, y, plus(z, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)

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

QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(0, y1, y2) → IF(true, 0, y1, plus(y2, s(0)))
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
minus(x, 0) → x
zero(s(x)) → false
zero(0) → true
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), x1)
zero(s(x0))
zero(0)

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.

zero(s(x0))
zero(0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

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

IF(false, x, s(y), z) → QUOT(minus(x, s(y)), s(y), z)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))
IF(false, 0, s(y1), y2) → QUOT(0, s(y1), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

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

IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)
QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule QUOT(s(x0), y1, y2) → IF(false, s(x0), y1, plus(y2, s(0))) we obtained the following new rules:

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

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

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))
IF(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), 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(false, s(x0), s(x1), y2) → QUOT(minus(x0, x1), s(x1), y2)

The remaining pairs can at least be oriented weakly.

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(IF(x₁, x₂, x₃, x₄)) = x₂ + x₃
POL(QUOT(x₁, x₂, x₃)) = x₁ + x₂
POL(false) = 0
POL(minus(x₁, x₂)) = x₁
POL(plus(x₁, x₂)) = 0
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ QDPOrderProof

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

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

QUOT(s(x0), s(z1), z2) → IF(false, s(x0), s(z1), plus(z2, s(0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
plus(0, x0)
plus(s(x0), 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.