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

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

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

Q is empty.

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

QUOT(x, y) → DIV(x, y, 0)
IF(true, true, x, y, z) → ID_INC(z)
IF(true, true, x, y, z) → MINUS(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF(true, true, x, y, z) → DIV(minus(x, y), y, id_inc(z))
DIV(x, y, z) → GE(y, s(0))
DIV(x, y, z) → GE(x, y)
GE(s(x), s(y)) → GE(x, y)
DIV(x, y, z) → IF(ge(y, s(0)), ge(x, y), x, y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

QUOT(x, y) → DIV(x, y, 0)
IF(true, true, x, y, z) → ID_INC(z)
IF(true, true, x, y, z) → MINUS(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
IF(true, true, x, y, z) → DIV(minus(x, y), y, id_inc(z))
DIV(x, y, z) → GE(y, s(0))
DIV(x, y, z) → GE(x, y)
GE(s(x), s(y)) → GE(x, y)
DIV(x, y, z) → IF(ge(y, s(0)), ge(x, y), x, y, z)

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

Q is empty.
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

  ↳ 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:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ UsableRulesProof

              ↳ 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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

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

The TRS R consists of the following rules:

ge(x, 0) → true
ge(0, s(y)) → false
ge(s(x), s(y)) → ge(x, y)
minus(x, 0) → x
minus(0, y) → 0
minus(s(x), s(y)) → minus(x, y)
id_inc(x) → x
id_inc(x) → s(x)
quot(x, y) → div(x, y, 0)
div(x, y, z) → if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z) → div_by_zero
if(true, false, x, y, z) → z
if(true, true, x, y, z) → div(minus(x, y), y, id_inc(z))

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