Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof

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

DX(times(ALPHA, BETA)) → DX(ALPHA)
DX(plus(ALPHA, BETA)) → DX(BETA)
DX(div(ALPHA, BETA)) → DX(ALPHA)
DX(neg(ALPHA)) → DX(ALPHA)
DX(times(ALPHA, BETA)) → DX(BETA)
DX(plus(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(BETA)
DX(exp(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(BETA)
DX(ln(ALPHA)) → DX(ALPHA)
DX(div(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.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ UsableRulesProof
        ↳ QDP
          ↳ QDPSizeChangeProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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: