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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 DependencyGraphProof (⇔)
    ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 PisEmptyProof (⇔)
    ↳15 TRUE
  ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE

(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) 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)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHOOSE(x1, x2, x3, x4)  =  CHOOSE(x2)
cons(x1, x2)  =  cons(x1, x2)
0  =  0
s(x1)  =  s
INSERT(x1, x2)  =  INSERT(x2)

Recursive Path Order [RPO].
Precedence:
s > [CHOOSE1, cons2, 0, INSERT1]


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

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.

(10) DependencyGraphProof (EQUIVALENT transformation)

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

(11) Obligation:

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

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.

(12) 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), s(y), s(z)) → CHOOSE(x, cons(v, w), y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
CHOOSE(x1, x2, x3, x4)  =  CHOOSE(x4)
cons(x1, x2)  =  cons
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
s1 > [CHOOSE1, cons]


The following usable rules [FROCOS05] were oriented: none

(13) 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)

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.

(14) PisEmptyProof (EQUIVALENT transformation)

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

(15) TRUE

(16) 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.

(17) 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: Combined order from the following AFS and order.
SORT(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) 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)

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.

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE