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:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

Q restricted rewrite system:
The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))


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

SORT1(cons2(x, y)) -> INSERT2(x, sort1(y))
INSERT2(x, cons2(v, w)) -> CHOOSE4(x, cons2(v, w), x, v)
CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)
CHOOSE4(x, cons2(v, w), 0, s1(z)) -> INSERT2(x, w)
SORT1(cons2(x, y)) -> SORT1(y)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

SORT1(cons2(x, y)) -> INSERT2(x, sort1(y))
INSERT2(x, cons2(v, w)) -> CHOOSE4(x, cons2(v, w), x, v)
CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)
CHOOSE4(x, cons2(v, w), 0, s1(z)) -> INSERT2(x, w)
SORT1(cons2(x, y)) -> SORT1(y)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
              ↳ QDP

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

INSERT2(x, cons2(v, w)) -> CHOOSE4(x, cons2(v, w), x, v)
CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)
CHOOSE4(x, cons2(v, w), 0, s1(z)) -> INSERT2(x, w)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHOOSE4(x, cons2(v, w), 0, s1(z)) -> INSERT2(x, w)
Used argument filtering: INSERT2(x1, x2)  =  x2
cons2(x1, x2)  =  cons1(x2)
CHOOSE4(x1, x2, x3, x4)  =  x2
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ DependencyGraphProof
              ↳ QDP

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

INSERT2(x, cons2(v, w)) -> CHOOSE4(x, cons2(v, w), x, v)
CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
              ↳ QDP

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

CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

CHOOSE4(x, cons2(v, w), s1(y), s1(z)) -> CHOOSE4(x, cons2(v, w), y, z)
Used argument filtering: CHOOSE4(x1, x2, x3, x4)  =  x4
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ DependencyGraphProof
                      ↳ QDP
                        ↳ QDPAfsSolverProof
                          ↳ QDP
                            ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(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
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof

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

SORT1(cons2(x, y)) -> SORT1(y)

The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

SORT1(cons2(x, y)) -> SORT1(y)
Used argument filtering: SORT1(x1)  =  x1
cons2(x1, x2)  =  cons1(x2)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPAfsSolverProof
                  ↳ QDP
                    ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

sort1(nil) -> nil
sort1(cons2(x, y)) -> insert2(x, sort1(y))
insert2(x, nil) -> cons2(x, nil)
insert2(x, cons2(v, w)) -> choose4(x, cons2(v, w), x, v)
choose4(x, cons2(v, w), y, 0) -> cons2(x, cons2(v, w))
choose4(x, cons2(v, w), 0, s1(z)) -> cons2(v, insert2(x, w))
choose4(x, cons2(v, w), s1(y), s1(z)) -> choose4(x, cons2(v, w), y, z)

The set Q consists of the following terms:

sort1(nil)
sort1(cons2(x0, x1))
insert2(x0, nil)
insert2(x0, cons2(x1, x2))
choose4(x0, cons2(x1, x2), x3, 0)
choose4(x0, cons2(x1, x2), 0, s1(x3))
choose4(x0, cons2(x1, x2), s1(x3), s1(x4))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.