Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 PisEmptyProof (⇔)
↳6 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

DX(plus(ALPHA, BETA)) → DX(ALPHA)
DX(plus(ALPHA, BETA)) → DX(BETA)
DX(times(ALPHA, BETA)) → DX(ALPHA)
DX(times(ALPHA, BETA)) → DX(BETA)
DX(minus(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(BETA)
DX(neg(ALPHA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(BETA)
DX(ln(ALPHA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(ALPHA)
DX(exp(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.

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


DX(plus(ALPHA, BETA)) → DX(ALPHA)
DX(plus(ALPHA, BETA)) → DX(BETA)
DX(times(ALPHA, BETA)) → DX(ALPHA)
DX(times(ALPHA, BETA)) → DX(BETA)
DX(minus(ALPHA, BETA)) → DX(ALPHA)
DX(minus(ALPHA, BETA)) → DX(BETA)
DX(neg(ALPHA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(ALPHA)
DX(div(ALPHA, BETA)) → DX(BETA)
DX(ln(ALPHA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(ALPHA)
DX(exp(ALPHA, BETA)) → DX(BETA)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
DX(x1)  =  x1
plus(x1, x2)  =  plus(x1, x2)
times(x1, x2)  =  times(x1, x2)
minus(x1, x2)  =  minus(x1, x2)
neg(x1)  =  neg(x1)
div(x1, x2)  =  div(x1, x2)
ln(x1)  =  ln(x1)
exp(x1, x2)  =  exp(x1, x2)
dx(x1)  =  dx(x1)
one  =  one
a  =  a
zero  =  zero
two  =  two

Lexicographic path order with status [LPO].
Precedence:
dx1 > neg1 > div2
dx1 > ln1 > div2
dx1 > exp2 > plus2 > div2
dx1 > exp2 > times2 > div2
dx1 > exp2 > minus2 > div2
dx1 > exp2 > one > div2
dx1 > zero > div2
dx1 > two > div2
a > zero > div2

Status:
plus2: [1,2]
times2: [1,2]
minus2: [1,2]
neg1: [1]
div2: [1,2]
ln1: [1]
exp2: [1,2]
dx1: [1]
one: []
a: []
zero: []
two: []

The following usable rules [FROCOS05] were oriented:

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))))

(4) Obligation:

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.

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE