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 [19] 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 [1,15] we result in the following initial DP problem:
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)
SORT(cons(x, y)) → SORT(y)
SORT(cons(x, y)) → INSERT(x, sort(y))
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.

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

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)
SORT(cons(x, y)) → SORT(y)
SORT(cons(x, y)) → INSERT(x, sort(y))
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 [15,17,22] contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
              ↳ QDP

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
              ↳ QDP

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)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof
              ↳ QDP

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)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof

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.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof

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

SORT(cons(x, y)) → SORT(y)

R is empty.
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 the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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))



↳ QTRS
  ↳ Overlay + Local Confluence
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ UsableRulesProof
                  ↳ QDP
                    ↳ QReductionProof
                      ↳ QDP
                        ↳ QDPSizeChangeProof

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

SORT(cons(x, y)) → SORT(y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: