Termination w.r.t. Q of the following Term Rewriting System could be proven:

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.


↳ QTRS
  ↳ Overlay + Local Confluence

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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof

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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

SORT(cons(x, y)) → INSERT(x, sort(y))
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)
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)
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.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof

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

SORT(cons(x, y)) → INSERT(x, sort(y))
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)
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)
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.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof

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

SORT(cons(x, y)) → INSERT(x, sort(y))
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)
SORT(cons(x, y)) → SORT(y)
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)

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.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                  ↳ QDP

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)
CHOOSE(x, cons(v, w), 0, s(z)) → INSERT(x, w)

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.
We use the reduction pair processor [13].


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.

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)
Used ordering: Combined order from the following AFS and order.
INSERT(x1, x2)  =  x2
cons(x1, x2)  =  cons(x2)
CHOOSE(x1, x2, x3, x4)  =  x2
s(x1)  =  x1
0  =  0

Lexicographic path order with status [19].
Precedence:
trivial

Status:
0: multiset
cons1: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                  ↳ QDP

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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof
                  ↳ QDP

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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
CHOOSE(x1, x2, x3, x4)  =  CHOOSE(x3)
cons(x1, x2)  =  cons(x1, x2)
s(x1)  =  s(x1)

Lexicographic path order with status [19].
Precedence:
CHOOSE1 > cons2
s1 > cons2

Status:
CHOOSE1: [1]
s1: multiset
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ DependencyGraphProof
                          ↳ QDP
                            ↳ QDPOrderProof
                              ↳ QDP
                                ↳ PisEmptyProof
                  ↳ QDP

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

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof

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.
We use the reduction pair processor [13].


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.
none
Used ordering: Combined order from the following AFS and order.
SORT(x1)  =  SORT(x1)
cons(x1, x2)  =  cons(x1, x2)

Lexicographic path order with status [19].
Precedence:
cons2 > SORT1

Status:
SORT1: [1]
cons2: multiset

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ EdgeDeletionProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ AND
                  ↳ QDP
                  ↳ QDP
                    ↳ QDPOrderProof
                      ↳ QDP
                        ↳ PisEmptyProof

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