Termination w.r.t. Q proof of TRS/Rubiopolo2.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:

dx₁(X) -> one
dx₁(a) -> zero
dx₁(plus₂(ALPHA, BETA)) -> plus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(times₂(ALPHA, BETA)) -> plus₂(times₂(BETA, dx₁(ALPHA)), times₂(ALPHA, dx₁(BETA)))
dx₁(minus₂(ALPHA, BETA)) -> minus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(neg₁(ALPHA)) -> neg₁(dx₁(ALPHA))
dx₁(div₂(ALPHA, BETA)) -> minus₂(div₂(dx₁(ALPHA), BETA), times₂(ALPHA, div₂(dx₁(BETA), exp₂(BETA, two))))
dx₁(ln₁(ALPHA)) -> div₂(dx₁(ALPHA), ALPHA)
dx₁(exp₂(ALPHA, BETA)) -> plus₂(times₂(BETA, times₂(exp₂(ALPHA, minus₂(BETA, one)), dx₁(ALPHA))), times₂(exp₂(ALPHA, BETA), times₂(ln₁(ALPHA), dx₁(BETA))))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

dx₁(X) -> one
dx₁(a) -> zero
dx₁(plus₂(ALPHA, BETA)) -> plus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(times₂(ALPHA, BETA)) -> plus₂(times₂(BETA, dx₁(ALPHA)), times₂(ALPHA, dx₁(BETA)))
dx₁(minus₂(ALPHA, BETA)) -> minus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(neg₁(ALPHA)) -> neg₁(dx₁(ALPHA))
dx₁(div₂(ALPHA, BETA)) -> minus₂(div₂(dx₁(ALPHA), BETA), times₂(ALPHA, div₂(dx₁(BETA), exp₂(BETA, two))))
dx₁(ln₁(ALPHA)) -> div₂(dx₁(ALPHA), ALPHA)
dx₁(exp₂(ALPHA, BETA)) -> plus₂(times₂(BETA, times₂(exp₂(ALPHA, minus₂(BETA, one)), dx₁(ALPHA))), times₂(exp₂(ALPHA, BETA), times₂(ln₁(ALPHA), dx₁(BETA))))

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

DX₁(exp₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)
DX₁(exp₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(neg₁(ALPHA)) -> DX₁(ALPHA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(ALPHA)

The TRS R consists of the following rules:

dx₁(X) -> one
dx₁(a) -> zero
dx₁(plus₂(ALPHA, BETA)) -> plus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(times₂(ALPHA, BETA)) -> plus₂(times₂(BETA, dx₁(ALPHA)), times₂(ALPHA, dx₁(BETA)))
dx₁(minus₂(ALPHA, BETA)) -> minus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(neg₁(ALPHA)) -> neg₁(dx₁(ALPHA))
dx₁(div₂(ALPHA, BETA)) -> minus₂(div₂(dx₁(ALPHA), BETA), times₂(ALPHA, div₂(dx₁(BETA), exp₂(BETA, two))))
dx₁(ln₁(ALPHA)) -> div₂(dx₁(ALPHA), ALPHA)
dx₁(exp₂(ALPHA, BETA)) -> plus₂(times₂(BETA, times₂(exp₂(ALPHA, minus₂(BETA, one)), dx₁(ALPHA))), times₂(exp₂(ALPHA, BETA), times₂(ln₁(ALPHA), dx₁(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:

DX₁(exp₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)
DX₁(exp₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(neg₁(ALPHA)) -> DX₁(ALPHA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(ALPHA)

The TRS R consists of the following rules:

dx₁(X) -> one
dx₁(a) -> zero
dx₁(plus₂(ALPHA, BETA)) -> plus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(times₂(ALPHA, BETA)) -> plus₂(times₂(BETA, dx₁(ALPHA)), times₂(ALPHA, dx₁(BETA)))
dx₁(minus₂(ALPHA, BETA)) -> minus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(neg₁(ALPHA)) -> neg₁(dx₁(ALPHA))
dx₁(div₂(ALPHA, BETA)) -> minus₂(div₂(dx₁(ALPHA), BETA), times₂(ALPHA, div₂(dx₁(BETA), exp₂(BETA, two))))
dx₁(ln₁(ALPHA)) -> div₂(dx₁(ALPHA), ALPHA)
dx₁(exp₂(ALPHA, BETA)) -> plus₂(times₂(BETA, times₂(exp₂(ALPHA, minus₂(BETA, one)), dx₁(ALPHA))), times₂(exp₂(ALPHA, BETA), times₂(ln₁(ALPHA), dx₁(BETA))))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

DX₁(exp₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)
DX₁(exp₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(neg₁(ALPHA)) -> DX₁(ALPHA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(times₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(div₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(plus₂(ALPHA, BETA)) -> DX₁(ALPHA)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DX₁(x₁)) = 3·x₁
POL(div₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(exp₂(x₁, x₂)) = 3 + x₁ + 2·x₂
POL(ln₁(x₁)) = 3 + 2·x₁
POL(minus₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(neg₁(x₁)) = 3 + 2·x₁
POL(plus₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂
POL(times₂(x₁, x₂)) = 3 + 2·x₁ + 2·x₂

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ PisEmptyProof

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

dx₁(X) -> one
dx₁(a) -> zero
dx₁(plus₂(ALPHA, BETA)) -> plus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(times₂(ALPHA, BETA)) -> plus₂(times₂(BETA, dx₁(ALPHA)), times₂(ALPHA, dx₁(BETA)))
dx₁(minus₂(ALPHA, BETA)) -> minus₂(dx₁(ALPHA), dx₁(BETA))
dx₁(neg₁(ALPHA)) -> neg₁(dx₁(ALPHA))
dx₁(div₂(ALPHA, BETA)) -> minus₂(div₂(dx₁(ALPHA), BETA), times₂(ALPHA, div₂(dx₁(BETA), exp₂(BETA, two))))
dx₁(ln₁(ALPHA)) -> div₂(dx₁(ALPHA), ALPHA)
dx₁(exp₂(ALPHA, BETA)) -> plus₂(times₂(BETA, times₂(exp₂(ALPHA, minus₂(BETA, one)), dx₁(ALPHA))), times₂(exp₂(ALPHA, BETA), times₂(ln₁(ALPHA), dx₁(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.