Termination w.r.t. Q proof of /tmp/tmplSMl9x/polo2.trs

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

      ↳ QDPOrderProof

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 use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

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 remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ln(x₁)) = 4 + (4)x_1
POL(plus(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(exp(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(DX(x₁)) = (4)x_1
POL(div(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(minus(x₁, x₂)) = 4 + (4)x_1 + (2)x_2
POL(times(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(neg(x₁)) = 4 + (4)x_1

The value of delta used in the strict ordering is 16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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