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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
ln(x1)  =  ln(x1)
neg(x1)  =  neg(x1)
div(x1, x2)  =  div(x1, x2)
exp(x1, x2)  =  exp(x1, x2)
minus(x1, x2)  =  minus(x1, x2)
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• DX(plus(ALPHA, BETA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(plus(ALPHA, BETA)) → DX(BETA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(times(ALPHA, BETA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(times(ALPHA, BETA)) → DX(BETA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(minus(ALPHA, BETA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(minus(ALPHA, BETA)) → DX(BETA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(neg(ALPHA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(div(ALPHA, BETA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(div(ALPHA, BETA)) → DX(BETA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(ln(ALPHA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(exp(ALPHA, BETA)) → DX(ALPHA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• DX(exp(ALPHA, BETA)) → DX(BETA) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(4) TRUE