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:

dx(X) → one
dx(a) → zero
dx(plus(ALPHA, BETA)) → plus(dx(ALPHA), dx(BETA))
dx(times(ALPHA, BETA)) → plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA)))
dx(minus(ALPHA, BETA)) → minus(dx(ALPHA), dx(BETA))
dx(neg(ALPHA)) → neg(dx(ALPHA))
dx(div(ALPHA, BETA)) → minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two))))
dx(ln(ALPHA)) → div(dx(ALPHA), ALPHA)
dx(exp(ALPHA, BETA)) → plus(times(BETA, times(exp(ALPHA, minus(BETA, one)), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA))))

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:

DX(plus(ALPHA, BETA)) → DX(ALPHA)
DX(plus(ALPHA, BETA)) → DX(BETA)
DX(times(ALPHA, BETA)) → DX(ALPHA)
DX(times(ALPHA, BETA)) → DX(BETA)
DX(minus(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(BETA)
DX(neg(ALPHA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(BETA)
DX(ln(ALPHA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(BETA)

The TRS R consists of the following rules:

dx(X) → one
dx(a) → zero
dx(plus(ALPHA, BETA)) → plus(dx(ALPHA), dx(BETA))
dx(times(ALPHA, BETA)) → plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA)))
dx(minus(ALPHA, BETA)) → minus(dx(ALPHA), dx(BETA))
dx(neg(ALPHA)) → neg(dx(ALPHA))
dx(div(ALPHA, BETA)) → minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two))))
dx(ln(ALPHA)) → div(dx(ALPHA), ALPHA)
dx(exp(ALPHA, BETA)) → plus(times(BETA, times(exp(ALPHA, minus(BETA, one)), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA))))

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.

DX(plus(ALPHA, BETA)) → DX(ALPHA)
DX(plus(ALPHA, BETA)) → DX(BETA)
DX(times(ALPHA, BETA)) → DX(ALPHA)
DX(times(ALPHA, BETA)) → DX(BETA)
DX(minus(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(BETA)
DX(div(ALPHA, BETA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(BETA)
DX(ln(ALPHA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(BETA)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DX(x1)  =  DX(x1)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)
minus(x1, x2)  =  minus(x1, x2)
neg(x1)  =  x1
div(x1, x2)  =  div(x1, x2)
ln(x1)  =  ln(x1)
exp(x1, x2)  =  exp(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:

plus₂ > DX₁
ln₁ > DX₁

Status:

ln₁: multiset
plus₂: [1,2]
exp₂: [1,2]
div₂: multiset
minus₂: [2,1]
DX₁: multiset
times₂: [1,2]

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

DX(neg(ALPHA)) → DX(ALPHA)

The TRS R consists of the following rules:

dx(X) → one
dx(a) → zero
dx(plus(ALPHA, BETA)) → plus(dx(ALPHA), dx(BETA))
dx(times(ALPHA, BETA)) → plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA)))
dx(minus(ALPHA, BETA)) → minus(dx(ALPHA), dx(BETA))
dx(neg(ALPHA)) → neg(dx(ALPHA))
dx(div(ALPHA, BETA)) → minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two))))
dx(ln(ALPHA)) → div(dx(ALPHA), ALPHA)
dx(exp(ALPHA, BETA)) → plus(times(BETA, times(exp(ALPHA, minus(BETA, one)), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA))))

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.

DX(neg(ALPHA)) → DX(ALPHA)

The remaining pairs can at least be oriented weakly.
Used ordering: Recursive path order with status [RPO].
Quasi-Precedence:

neg₁ > DX₁

Status:

DX₁: multiset
neg₁: multiset

The following usable rules [FROCOS05] were oriented: none

(6) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

dx(X) → one
dx(a) → zero
dx(plus(ALPHA, BETA)) → plus(dx(ALPHA), dx(BETA))
dx(times(ALPHA, BETA)) → plus(times(BETA, dx(ALPHA)), times(ALPHA, dx(BETA)))
dx(minus(ALPHA, BETA)) → minus(dx(ALPHA), dx(BETA))
dx(neg(ALPHA)) → neg(dx(ALPHA))
dx(div(ALPHA, BETA)) → minus(div(dx(ALPHA), BETA), times(ALPHA, div(dx(BETA), exp(BETA, two))))
dx(ln(ALPHA)) → div(dx(ALPHA), ALPHA)
dx(exp(ALPHA, BETA)) → plus(times(BETA, times(exp(ALPHA, minus(BETA, one)), dx(ALPHA))), times(exp(ALPHA, BETA), times(ln(ALPHA), dx(BETA))))

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