(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: Combined order from the following AFS and order.
MERGE(
x1,
x2) =
MERGE(
x1)
++(
x1,
x2) =
++(
x1,
x2)
u =
u
v =
v
Recursive Path Order [RPO].
Precedence:
++2 > u > MERGE1 > v
The following usable rules [FROCOS05] were oriented:
none
(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