Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(0) Obligation:

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.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(4) Obligation:

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.

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


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.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
MERGE2 > [++2, v] > u
merge2 > [++2, v] > u


The following usable rules [FROCOS05] were oriented:

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))

(6) Obligation:

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.

(7) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(8) TRUE