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:

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

Q is empty.

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

MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(++2(x, y), v)
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))

The TRS R consists of the following rules:

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(++2(x, y), v)
MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))

The TRS R consists of the following rules:

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPAfsSolverProof

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

MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))

The TRS R consists of the following rules:

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

Q is empty.
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.

MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
Used argument filtering: MERGE2(x1, x2)  =  x1
++2(x1, x2)  =  ++1(x2)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPAfsSolverProof
QDP
              ↳ PisEmptyProof

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

merge2(x, nil) -> x
merge2(nil, y) -> y
merge2(++2(x, y), ++2(u, v)) -> ++2(x, merge2(y, ++2(u, v)))
merge2(++2(x, y), ++2(u, v)) -> ++2(u, merge2(++2(x, y), v))

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.