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.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


MERGE2(++2(x, y), ++2(u, v)) -> MERGE2(y, ++2(u, v))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(++2(x1, x2)) = 1 + x2   
POL(MERGE2(x1, x2)) = 3·x1   
POL(u) = 0   
POL(v) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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.