(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

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.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


D1(+(x, y)) → D1(x)
D1(+(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
D1(minus(x)) → D1(x)
D1(div(x, y)) → D1(x)
D1(div(x, y)) → D1(y)
D1(ln(x)) → D1(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*(x1, x2)) = 1 + x1 + x2   
POL(+(x1, x2)) = 1 + x1 + x2   
POL(-(x1, x2)) = 1 + x1 + x2   
POL(D1(x1)) = x1   
POL(div(x1, x2)) = 1 + x1 + x2   
POL(ln(x1)) = 1 + x1   
POL(minus(x1)) = 1 + x1   
POL(pow(x1, x2)) = x1 + x2   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

D1(pow(x, y)) → D1(x)
D1(pow(x, y)) → D1(y)

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


D1(pow(x, y)) → D1(x)
D1(pow(x, y)) → D1(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(D1(x1)) = x1   
POL(pow(x1, x2)) = 1 + x1 + x2   

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE