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

(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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
*(x1, x2)  =  *(x1, x2)
+(x1, x2)  =  +(x1, x2)
-(x1, x2)  =  -(x1, x2)

From the DPs we obtained the following set of size-change graphs:

• D¹(+(x, y)) → D¹(x) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• D¹(+(x, y)) → D¹(y) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• D¹(*(x, y)) → D¹(x) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• D¹(*(x, y)) → D¹(y) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• D¹(-(x, y)) → D¹(x) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

• D¹(-(x, y)) → D¹(y) (allowed arguments on rhs = {1})
  The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(4) TRUE