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(div2(ALPHA, BETA)) -> DX1(BETA)
DX1(minus2(ALPHA, BETA)) -> DX1(BETA)
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.

DX1(times2(ALPHA, BETA)) -> DX1(BETA)
DX1(times2(ALPHA, BETA)) -> DX1(ALPHA)
Used ordering: Polynomial interpretation [21]:

POL(DX1(x1)) = x1   
POL(div2(x1, x2)) = 1 + x1 + x2   
POL(exp2(x1, x2)) = 1 + x1 + x2   
POL(ln1(x1)) = 1 + x1   
POL(minus2(x1, x2)) = 1 + x1 + x2   
POL(neg1(x1)) = 1 + x1   
POL(plus2(x1, x2)) = 1 + x1 + x2   
POL(times2(x1, x2)) = x1 + x2   

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(times2(ALPHA, BETA)) -> DX1(BETA)
DX1(times2(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(times2(ALPHA, BETA)) -> DX1(BETA)
DX1(times2(ALPHA, BETA)) -> DX1(ALPHA)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(DX1(x1)) = x1   
POL(times2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ 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.