Termination w.r.t. Q proof of /tmp/tmpy26MYY/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

  ↳ 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)))

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:

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)))

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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)))

Q is empty.
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.

D¹(div(x, y)) → D¹(x)
D¹(div(x, y)) → D¹(y)

The remaining pairs can at least be oriented weakly.

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(pow(x, y)) → D¹(y)
D¹(-(x, y)) → D¹(x)
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)

Used ordering: Polynomial interpretation [25,35]:

POL(ln(x₁)) = (4)x_1
POL(-(x₁, x₂)) = (3/2)x_1 + (9/4)x_2
POL(div(x₁, x₂)) = 1/4 + (2)x_1 + x_2
POL(minus(x₁)) = (4)x_1
POL(*(x₁, x₂)) = (2)x_1 + (9/4)x_2
POL(D¹(x₁)) = x_1
POL(pow(x₁, x₂)) = (2)x_1 + x_2
POL(+(x₁, x₂)) = x_1 + (7/4)x_2

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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(-(x, y)) → D¹(x)
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)

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.
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.

D¹(+(x, y)) → D¹(y)
D¹(*(x, y)) → D¹(y)
D¹(pow(x, y)) → D¹(y)
D¹(-(x, y)) → 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)

The remaining pairs can at least be oriented weakly.

D¹(minus(x)) → D¹(x)

Used ordering: Polynomial interpretation [25,35]:

POL(ln(x₁)) = 4 + (11/4)x_1
POL(-(x₁, x₂)) = 1/4 + (2)x_1 + (9/4)x_2
POL(minus(x₁)) = (4)x_1
POL(*(x₁, x₂)) = 3 + (4)x_1 + (4)x_2
POL(D¹(x₁)) = (2)x_1
POL(pow(x₁, x₂)) = 1/4 + (5/2)x_1 + (2)x_2
POL(+(x₁, x₂)) = 1/2 + (4)x_1 + (4)x_2

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

D¹(minus(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)))

Q is empty.
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.

D¹(minus(x)) → D¹(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(minus(x₁)) = 1/4 + (7/2)x_1
POL(D¹(x₁)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

                ↳ QDP

                  ↳ PisEmptyProof

Q DP problem:
P is empty.
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.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.