Termination w.r.t. Q of the given QTRS could be proven:

(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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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)))

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))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) 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)))

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.

(5) 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)
D¹(pow(x, y)) → D¹(x)
D¹(pow(x, y)) → D¹(y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
D¹(x1)  =  x1
+(x1, x2)  =  +(x1, x2)
*(x1, x2)  =  *(x1, x2)
-(x1, x2)  =  -(x1, x2)
minus(x1)  =  minus(x1)
div(x1, x2)  =  div(x1, x2)
ln(x1)  =  ln(x1)
pow(x1, x2)  =  pow(x1, x2)

Lexicographic path order with status [LPO].
Precedence:

trivial

Status:

ln₁: [1]
-₂: [1,2]
div₂: [1,2]
minus₁: [1]
*₂: [1,2]
pow₂: [1,2]
+₂: [1,2]

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)))

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.

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE