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

(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) 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:

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)

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 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) 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)

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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 remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
CHOOSE(x0, x1, x2, x3, x4)  =  CHOOSE(x1, x2, x4)
INSERT(x0, x1, x2)  =  INSERT(x0, x1, x2)

Tags:
CHOOSE has argument tags [0,0,4,1,0] and root tag 0
INSERT has argument tags [2,0,5] and root tag 1

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
CHOOSE(x1, x2, x3, x4)  =  CHOOSE(x1, x2, x3, x4)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
s(x1)  =  s(x1)
INSERT(x1, x2)  =  INSERT

Lexicographic path order with status [LPO].
Quasi-Precedence:

0 > [CHOOSE₄, cons₂]
s₁ > INSERT > [CHOOSE₄, cons₂]

Status:

CHOOSE₄: [2,4,3,1]
cons₂: [2,1]
0: []
s₁: [1]
INSERT: []

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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)

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SORT(cons(x, y)) → SORT(y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SORT(x0, x1)  =  SORT(x0, x1)

Tags:
SORT has argument tags [1,0] and root tag 0

Comparison: MS
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
SORT(x1)  =  SORT
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [LPO].
Quasi-Precedence:

trivial

Status:

SORT: []
cons₂: [2,1]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE