Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar14/RubioSLASHpolo2.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 strictly oriented and are deleted.

DX₁(exp₂(ALPHA, BETA)) -> DX₁(ALPHA)
DX₁(exp₂(ALPHA, BETA)) -> DX₁(BETA)
DX₁(minus₂(ALPHA, BETA)) -> 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 by weakly be oriented.

DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)
DX₁(neg₁(ALPHA)) -> DX₁(ALPHA)

Used ordering: Combined order from the following AFS and order.
DX₁(x1) = x1
exp₂(x1, x2) = exp₂(x1, x2)
ln₁(x1) = x1
minus₂(x1, x2) = minus₂(x1, x2)
neg₁(x1) = x1
times₂(x1, x2) = times₂(x1, x2)
div₂(x1, x2) = div₂(x1, x2)
plus₂(x1, x2) = plus₂(x1, x2)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)
DX₁(neg₁(ALPHA)) -> 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 strictly oriented and are deleted.

DX₁(neg₁(ALPHA)) -> DX₁(ALPHA)

The remaining pairs can at least by weakly be oriented.

DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)

Used ordering: Combined order from the following AFS and order.
DX₁(x1) = DX₁(x1)
ln₁(x1) = x1
neg₁(x1) = neg₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ QDPOrderProof

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

DX₁(ln₁(ALPHA)) -> 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 strictly oriented and are deleted.

DX₁(ln₁(ALPHA)) -> DX₁(ALPHA)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
DX₁(x1) = DX₁(x1)
ln₁(x1) = ln₁(x1)

Lexicographic Path Order [19].
Precedence:

ln₁ > DX₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ 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.