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:

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.


↳ QTRS
  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

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

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


D1(+(x, y)) → D1(y)
D1(-(x, y)) → D1(x)
D1(-(x, y)) → D1(y)
D1(+(x, y)) → D1(x)
The remaining pairs can at least be oriented weakly.

D1(*(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
Used ordering: Polynomial interpretation [25,35]:

POL(-(x1, x2)) = 1/2 + (4)x_1 + x_2   
POL(*(x1, x2)) = (5/2)x_1 + (5/4)x_2   
POL(D1(x1)) = (2)x_1   
POL(+(x1, x2)) = 1/4 + (5/2)x_1 + (7/2)x_2   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ QDPOrderProof

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

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

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


D1(*(x, y)) → D1(y)
D1(*(x, y)) → D1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(*(x1, x2)) = 1/4 + (5/2)x_1 + (5/2)x_2   
POL(D1(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

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