Termination w.r.t. Q proof of /tmp/tmpMEDAyo/modulo.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MOD(x, y) → MINUS(x, y)
MOD(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MOD(x, y) → MINUS(x, y)
MOD(x, y) → LE(y, x)
MINUS(s(x), s(y)) → MINUS(x, y)
MOD(x, y) → ISZERO(y)
MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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 3 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:

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MINUS(s(x), s(y)) → MINUS(x, 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:

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

                ↳ UsableRulesProof

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

LE(s(x), s(y)) → LE(x, y)

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

LE(s(x), s(y)) → LE(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:

MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
isZero(0) → true
isZero(s(x)) → false
mod(x, y) → if_mod(isZero(y), le(y, x), x, y, minus(x, y))
if_mod(true, b, x, y, z) → divByZeroError
if_mod(false, false, x, y, z) → x
if_mod(false, true, x, y, z) → mod(z, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))
mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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.

mod(x0, x1)
if_mod(true, x0, x1, x2, x3)
if_mod(false, false, x0, x1, x2)
if_mod(false, true, x0, 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:

MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(x, y) → IF_MOD(isZero(y), le(y, x), x, y, minus(x, y)) at position [0] we obtained the following new rules:

MOD(y0, 0) → IF_MOD(true, le(0, y0), y0, 0, minus(y0, 0))
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))

↳ 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:

MOD(y0, 0) → IF_MOD(true, le(0, y0), y0, 0, minus(y0, 0))
MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))

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:

MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

isZero(0) → true
isZero(s(x)) → false
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, x) → 0
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))

↳ 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:

MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, x0)
minus(x0, 0)
minus(0, x0)
minus(s(x0), s(x1))
isZero(0)
isZero(s(x0))

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.

isZero(0)
isZero(s(x0))

↳ 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:

MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(y0, s(x0)) → IF_MOD(false, le(s(x0), y0), y0, s(x0), minus(y0, s(x0))) at position [1] we obtained the following new rules:

MOD(0, s(x0)) → IF_MOD(false, false, 0, s(x0), minus(0, s(x0)))
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))

↳ 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:

MOD(0, s(x0)) → IF_MOD(false, false, 0, s(x0), minus(0, s(x0)))
IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

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

IF_MOD(false, true, x, y, z) → MOD(z, y)
MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0)))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MOD(s(x1), s(x0)) → IF_MOD(false, le(x0, x1), s(x1), s(x0), minus(s(x1), s(x0))) at position [4] we obtained the following new rules:

MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 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

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

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

MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, x, y, z) → MOD(z, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

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

IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))
IF_MOD(false, true, s(z0), s(z0), 0) → MOD(0, s(z0))

↳ 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

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

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

MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))
IF_MOD(false, true, s(z0), s(z0), 0) → MOD(0, s(z0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ ForwardInstantiation

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

MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)
IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF_MOD(false, true, s(z0), s(z1), y_1) → MOD(y_1, s(z1)) we obtained the following new rules:

IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))

↳ 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

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ ForwardInstantiation

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

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

IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))
MOD(s(y0), s(y0)) → IF_MOD(false, le(y0, y0), s(y0), s(y0), 0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ UsableRulesProof

                                  ↳ QDP

                                    ↳ QReductionProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ ForwardInstantiation

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

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

IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))
MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

MOD(s(x0), s(x1)) → IF_MOD(false, le(x1, x0), s(x0), s(x1), minus(x0, x1))

The remaining pairs can at least be oriented weakly.

IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))

Used ordering: Polynomial interpretation [25]:

POL(0) = 0
POL(IF_MOD(x₁, x₂, x₃, x₄, x₅)) = x₅
POL(MOD(x₁, x₂)) = x₁
POL(false) = 0
POL(le(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, x) → 0
minus(x, x) → 0
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)

↳ 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

                                                ↳ Narrowing

                                                  ↳ QDP

                                                    ↳ Instantiation

                                                      ↳ QDP

                                                        ↳ DependencyGraphProof

                                                          ↳ QDP

                                                            ↳ ForwardInstantiation

                                                              ↳ QDP

                                                                ↳ DependencyGraphProof

                                                                  ↳ QDP

                                                                    ↳ QDPOrderProof

                                                                      ↳ QDP

                                                                        ↳ DependencyGraphProof

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

IF_MOD(false, true, s(x0), s(x1), s(y_0)) → MOD(s(y_0), s(x1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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