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.

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
      ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

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)
Used argument filtering: 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)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DX1(neg1(ALPHA)) -> DX1(ALPHA)
Used argument filtering: DX1(x1)  =  x1
ln1(x1)  =  x1
neg1(x1)  =  neg1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

DX1(ln1(ALPHA)) -> DX1(ALPHA)
Used argument filtering: DX1(x1)  =  x1
ln1(x1)  =  ln1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPAfsSolverProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ 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.