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
  ↳ 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))

Q is empty.

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))

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:

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))

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


The following pairs can be oriented strictly and are deleted.


D11(-2(x, y)) -> D11(y)
D11(-2(x, y)) -> D11(x)
The remaining pairs can at least be oriented weakly.

D11(*2(x, y)) -> D11(x)
D11(*2(x, y)) -> D11(y)
D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = x1 + x2   
POL(+2(x1, x2)) = x1 + x2   
POL(-2(x1, x2)) = 1 + x1 + x2   
POL(D11(x1)) = x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ 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)

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


The following pairs can be oriented strictly and are deleted.


D11(*2(x, y)) -> D11(x)
D11(*2(x, y)) -> D11(y)
The remaining pairs can at least be oriented weakly.

D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
Used ordering: Polynomial interpretation [21]:

POL(*2(x1, x2)) = 1 + x1 + x2   
POL(+2(x1, x2)) = x1 + x2   
POL(D11(x1)) = x1   

The following usable rules [14] were oriented: none



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

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


The following pairs can be oriented strictly and are deleted.


D11(+2(x, y)) -> D11(x)
D11(+2(x, y)) -> D11(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(+2(x1, x2)) = 1 + x1 + x2   
POL(D11(x1)) = x1   

The following usable rules [14] were oriented: none



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

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.