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:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

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:

D1(+(x, y)) → D1(x)
D1(+(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)

The TRS R consists of the following rules:

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

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.


D1(+(x, y)) → D1(x)
D1(+(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
D1(*(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(*(x1, x2)) = 1 + x1 + x2   
POL(+(x1, x2)) = 1 + x1 + x2   
POL(-(x1, x2)) = 1 + x1 + x2   
POL(D1(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

D(t) → 1
D(constant) → 0
D(+(x, y)) → +(D(x), D(y))
D(*(x, y)) → +(*(y, D(x)), *(x, D(y)))
D(-(x, y)) → -(D(x), D(y))

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