(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)
Q is empty.
(1) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)
The set Q consists of the following terms:
sort(nil)
sort(cons(x0, x1))
insert(x0, nil)
insert(x0, cons(x1, x2))
choose(x0, cons(x1, x2), x3, 0)
choose(x0, cons(x1, x2), 0, s(x3))
choose(x0, cons(x1, x2), s(x3), s(x4))
(3) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(4) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SORT(cons(x, y)) → INSERT(x, sort(y))
SORT(cons(x, y)) → SORT(y)
INSERT(x, cons(v, w)) → CHOOSE(x, cons(v, w), x, v)
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)
CHOOSE(x, cons(v, w), s(y), s(z)) → CHOOSE(x, cons(v, w), y, z)
The TRS R consists of the following rules:
sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)
The set Q consists of the following terms:
sort(nil)
sort(cons(x0, x1))
insert(x0, nil)
insert(x0, cons(x1, x2))
choose(x0, cons(x1, x2), x3, 0)
choose(x0, cons(x1, x2), 0, s(x3))
choose(x0, cons(x1, x2), s(x3), s(x4))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)
INSERT(x, cons(v, w)) → CHOOSE(x, cons(v, w), x, v)
CHOOSE(x, cons(v, w), s(y), s(z)) → CHOOSE(x, cons(v, w), y, z)
The TRS R consists of the following rules:
sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)
The set Q consists of the following terms:
sort(nil)
sort(cons(x0, x1))
insert(x0, nil)
insert(x0, cons(x1, x2))
choose(x0, cons(x1, x2), x3, 0)
choose(x0, cons(x1, x2), 0, s(x3))
choose(x0, cons(x1, x2), s(x3), s(x4))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
SORT(cons(x, y)) → SORT(y)
The TRS R consists of the following rules:
sort(nil) → nil
sort(cons(x, y)) → insert(x, sort(y))
insert(x, nil) → cons(x, nil)
insert(x, cons(v, w)) → choose(x, cons(v, w), x, v)
choose(x, cons(v, w), y, 0) → cons(x, cons(v, w))
choose(x, cons(v, w), 0, s(z)) → cons(v, insert(x, w))
choose(x, cons(v, w), s(y), s(z)) → choose(x, cons(v, w), y, z)
The set Q consists of the following terms:
sort(nil)
sort(cons(x0, x1))
insert(x0, nil)
insert(x0, cons(x1, x2))
choose(x0, cons(x1, x2), x3, 0)
choose(x0, cons(x1, x2), 0, s(x3))
choose(x0, cons(x1, x2), s(x3), s(x4))
We have to consider all minimal (P,Q,R)-chains.