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
  ↳ Non-Overlap Check

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.

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:

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

The set Q consists of the following terms:

-2(-2(neg1(x0), neg1(x0)), -2(neg1(x1), neg1(x1)))


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))

The set Q consists of the following terms:

-2(-2(neg1(x0), neg1(x0)), -2(neg1(x1), neg1(x1)))

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

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

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))

The set Q consists of the following terms:

-2(-2(neg1(x0), neg1(x0)), -2(neg1(x1), neg1(x1)))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

-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)
Used argument filtering: -12(x1, x2)  =  x2
-2(x1, x2)  =  x2
neg1(x1)  =  neg1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPAfsSolverProof
            ↳ 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))

The set Q consists of the following terms:

-2(-2(neg1(x0), neg1(x0)), -2(neg1(x1), neg1(x1)))

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