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:

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, 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:

-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(-2(x, y), -2(x, y))
-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(x, y)

The TRS R consists of the following rules:

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, 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:

-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(-2(x, y), -2(x, y))
-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(x, y)

The TRS R consists of the following rules:

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, 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.


-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(-2(x, y), -2(x, y))
-12(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -12(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( -2(x1, x2) ) = 2x2


POL( neg1(x1) ) = 3x1 + 2


POL( -12(x1, x2) ) = max{0, x2 - 1}



The following usable rules [14] were oriented:

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, y))



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

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

-2(-2(neg1(x), neg1(x)), -2(neg1(y), neg1(y))) -> -2(-2(x, y), -2(x, 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.