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:

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(x, y), v))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(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:

MERGE(++(x, y), ++(u, v)) → MERGE(++(x, y), v)
MERGE(++(x, y), ++(u, v)) → MERGE(y, ++(u, v))

The TRS R consists of the following rules:

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(x, y), v))

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

MERGE(++(x, y), ++(u, v)) → MERGE(++(x, y), v)
MERGE(++(x, y), ++(u, v)) → MERGE(y, ++(u, v))

The TRS R consists of the following rules:

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(x, y), v))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

MERGE(++(x, y), ++(u, v)) → MERGE(++(x, y), v)
MERGE(++(x, y), ++(u, v)) → MERGE(y, ++(u, v))

The TRS R consists of the following rules:

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof

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

MERGE(++(x, y), ++(u, v)) → MERGE(y, ++(u, v))

The TRS R consists of the following rules:

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(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.


MERGE(++(x, y), ++(u, v)) → MERGE(y, ++(u, v))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MERGE(x1, x2)  =  x1
++(x1, x2)  =  ++(x1, x2)
u  =  u
v  =  v

Recursive Path Order [2].
Precedence:
u > ++2
v > ++2

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ QDP
                  ↳ PisEmptyProof

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

merge(x, nil) → x
merge(nil, y) → y
merge(++(x, y), ++(u, v)) → ++(x, merge(y, ++(u, v)))
merge(++(x, y), ++(u, v)) → ++(u, merge(++(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.