Termination w.r.t. Q proof of /tmp/tmp65cJCP/11.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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
D(minus(x0))
D(div(x0, x1))
D(ln(x0))
D(pow(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:

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(pow(x, y)) → D¹(y)
D¹(div(x, y)) → D¹(x)
D¹(-(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(minus(x)) → D¹(x)
D¹(-(x, y)) → D¹(y)
D¹(+(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(*(x, y)) → D¹(x)
D¹(ln(x)) → D¹(x)

The TRS R consists of the following rules:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
D(minus(x0))
D(div(x0, x1))
D(ln(x0))
D(pow(x0, x1))

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

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

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(pow(x, y)) → D¹(y)
D¹(div(x, y)) → D¹(x)
D¹(-(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(minus(x)) → D¹(x)
D¹(-(x, y)) → D¹(y)
D¹(+(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(*(x, y)) → D¹(x)
D¹(ln(x)) → D¹(x)

The TRS R consists of the following rules:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))
D(minus(x)) → minus(D(x))
D(div(x, y)) → -(div(D(x), y), div(*(x, D(y)), pow(y, 2)))
D(ln(x)) → div(D(x), x)
D(pow(x, y)) → +(*(*(y, pow(x, -(y, 1))), D(x)), *(*(pow(x, y), ln(x)), D(y)))

The set Q consists of the following terms:

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
D(minus(x0))
D(div(x0, x1))
D(ln(x0))
D(pow(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

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

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

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(pow(x, y)) → D¹(y)
D¹(div(x, y)) → D¹(x)
D¹(-(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(minus(x)) → D¹(x)
D¹(-(x, y)) → D¹(y)
D¹(+(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(*(x, y)) → D¹(x)
D¹(ln(x)) → D¹(x)

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

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
D(minus(x0))
D(div(x0, x1))
D(ln(x0))
D(pow(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.

D(t)
D(constant)
D(+(x0, x1))
D(*(x0, x1))
D(-(x0, x1))
D(minus(x0))
D(div(x0, x1))
D(ln(x0))
D(pow(x0, x1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ UsableRulesProof

            ↳ QDP

              ↳ QReductionProof

                ↳ QDP

                  ↳ QDPSizeChangeProof

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

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(div(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(y)
D¹(-(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)
D¹(minus(x)) → D¹(x)
D¹(+(x, y)) → D¹(x)
D¹(-(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(x)
D¹(ln(x)) → D¹(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:

D¹(+(x, y)) → D¹(y)
The graph contains the following edges 1 > 1
D¹(*(x, y)) → D¹(y)
The graph contains the following edges 1 > 1
D¹(pow(x, y)) → D¹(y)
The graph contains the following edges 1 > 1
D¹(div(x, y)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(-(x, y)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(div(x, y)) → D¹(y)
The graph contains the following edges 1 > 1
D¹(minus(x)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(-(x, y)) → D¹(y)
The graph contains the following edges 1 > 1
D¹(+(x, y)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(pow(x, y)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(*(x, y)) → D¹(x)
The graph contains the following edges 1 > 1
D¹(ln(x)) → D¹(x)
The graph contains the following edges 1 > 1