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:

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

The set Q consists of the following terms:

D1(t)
D1(constant)
D1(+2(x0, x1))
D1(*2(x0, x1))
D1(-2(x0, x1))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

D11(*2(x, y)) -> D11(x)
D11(*2(x, y)) -> D11(y)
D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(x)

The TRS R consists of the following rules:

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

The set Q consists of the following terms:

D1(t)
D1(constant)
D1(+2(x0, x1))
D1(*2(x0, x1))
D1(-2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof

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

D11(*2(x, y)) -> D11(x)
D11(*2(x, y)) -> D11(y)
D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(x)

The TRS R consists of the following rules:

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

The set Q consists of the following terms:

D1(t)
D1(constant)
D1(+2(x0, x1))
D1(*2(x0, x1))
D1(-2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


D11(*2(x, y)) -> D11(x)
D11(*2(x, y)) -> D11(y)
D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
D11(x1)  =  D11(x1)
*2(x1, x2)  =  *2(x1, x2)
+2(x1, x2)  =  +2(x1, x2)
-2(x1, x2)  =  -2(x1, x2)

Lexicographic Path Order [19].
Precedence:
trivial


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ PisEmptyProof

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

D1(t) -> 1
D1(constant) -> 0
D1(+2(x, y)) -> +2(D1(x), D1(y))
D1(*2(x, y)) -> +2(*2(y, D1(x)), *2(x, D1(y)))
D1(-2(x, y)) -> -2(D1(x), D1(y))

The set Q consists of the following terms:

D1(t)
D1(constant)
D1(+2(x0, x1))
D1(*2(x0, x1))
D1(-2(x0, x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.