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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(6) 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)

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.

(7) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(8) 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)

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] 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(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(*(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(-(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(div(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(ln(x)) → 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(pow(x, y)) → D¹(y)
  The graph contains the following edges 1 > 1

(10) TRUE