Termination Proof using AProVE

Term Rewriting System R:
[x, y, v, w, z]
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)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

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)
INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

INSERT(x, cons(v, w)) -> CHOOSE(x, cons(v, w), x, v)

two new Dependency Pairs are created:

INSERT(0, cons(s(z''), w'')) -> CHOOSE(0, cons(s(z''), w''), 0, s(z''))
INSERT(s(y''), cons(s(z''), w'')) -> CHOOSE(s(y''), cons(s(z''), w''), s(y''), s(z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(s(y''), cons(s(z''), w'')) -> CHOOSE(s(y''), cons(s(z''), w''), s(y''), s(z''))
INSERT(0, cons(s(z''), w'')) -> CHOOSE(0, cons(s(z''), w''), 0, s(z''))
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)

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

two new Dependency Pairs are created:

CHOOSE(0, cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(0, cons(s(z''''), w''''))
CHOOSE(s(y''''), cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(s(y''''), cons(s(z''''), w''''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(0, cons(s(z''), w'')) -> CHOOSE(0, cons(s(z''), w''), 0, s(z''))
CHOOSE(0, cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(0, cons(s(z''''), w''''))

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)

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

CHOOSE(0, cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(0, cons(s(z''''), w''''))

one new Dependency Pair is created:

CHOOSE(0, cons(s(z'), cons(s(z'''''), w''''')), 0, s(z')) -> INSERT(0, cons(s(z'''''), w'''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(0, cons(s(z'), cons(s(z'''''), w''''')), 0, s(z')) -> INSERT(0, cons(s(z'''''), w'''''))
INSERT(0, cons(s(z''), w'')) -> CHOOSE(0, cons(s(z''), w''), 0, s(z''))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

INSERT(0, cons(s(z''), w'')) -> CHOOSE(0, cons(s(z''), w''), 0, s(z''))

one new Dependency Pair is created:

INSERT(0, cons(s(z''''), cons(s(z'''''''), w'''''''))) -> CHOOSE(0, cons(s(z''''), cons(s(z'''''''), w''''''')), 0, s(z''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(0, cons(s(z''''), cons(s(z'''''''), w'''''''))) -> CHOOSE(0, cons(s(z''''), cons(s(z'''''''), w''''''')), 0, s(z''''))
CHOOSE(0, cons(s(z'), cons(s(z'''''), w''''')), 0, s(z')) -> INSERT(0, cons(s(z'''''), w'''''))

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

CHOOSE(0, cons(s(z'), cons(s(z'''''), w''''')), 0, s(z')) -> INSERT(0, cons(s(z'''''), w'''''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(CHOOSE(x₁, x₂, x₃, x₄)) = x₂

POL(0) = 0

POL(cons(x₁, x₂)) = x₁ + x₂

POL(INSERT(x₁, x₂)) = x₂

POL(s(x₁)) = 1

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 14

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

INSERT(0, cons(s(z''''), cons(s(z'''''''), w'''''''))) -> CHOOSE(0, cons(s(z''''), cons(s(z'''''''), w''''''')), 0, s(z''''))

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(s(y''''), cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(s(y''''), cons(s(z''''), w''''))
CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)
INSERT(s(y''), cons(s(z''), w'')) -> CHOOSE(s(y''), cons(s(z''), w''), s(y''), s(z''))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

CHOOSE(x, cons(v, w), s(y), s(z)) -> CHOOSE(x, cons(v, w), y, z)

two new Dependency Pairs are created:

CHOOSE(x'', cons(v'', w''), s(s(y'')), s(s(z''))) -> CHOOSE(x'', cons(v'', w''), s(y''), s(z''))
CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))
CHOOSE(x'', cons(v'', w''), s(s(y'')), s(s(z''))) -> CHOOSE(x'', cons(v'', w''), s(y''), s(z''))
INSERT(s(y''), cons(s(z''), w'')) -> CHOOSE(s(y''), cons(s(z''), w''), s(y''), s(z''))
CHOOSE(s(y''''), cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(s(y''''), cons(s(z''''), w''''))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

INSERT(s(y''), cons(s(z''), w'')) -> CHOOSE(s(y''), cons(s(z''), w''), s(y''), s(z''))

two new Dependency Pairs are created:

INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))
CHOOSE(x'', cons(v'', w''), s(s(y'')), s(s(z''))) -> CHOOSE(x'', cons(v'', w''), s(y''), s(z''))
INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(y''''), cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(s(y''''), cons(s(z''''), w''''))
CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

CHOOSE(s(y''''), cons(v, cons(s(z''''), w'''')), 0, s(z)) -> INSERT(s(y''''), cons(s(z''''), w''''))

two new Dependency Pairs are created:

CHOOSE(s(s(y'''''')), cons(v, cons(s(s(z'''''')), w'''''')), 0, s(z)) -> INSERT(s(s(y'''''')), cons(s(s(z'''''')), w''''''))
CHOOSE(s(0), cons(v, cons(s(s(z'''''')), cons(s(z''''''''''), w''''''''''))), 0, s(z)) -> INSERT(s(0), cons(s(s(z'''''')), cons(s(z''''''''''), w'''''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(s(0), cons(v, cons(s(s(z'''''')), cons(s(z''''''''''), w''''''''''))), 0, s(z)) -> INSERT(s(0), cons(s(s(z'''''')), cons(s(z''''''''''), w'''''''''')))
CHOOSE(x'', cons(v'', w''), s(s(y'')), s(s(z''))) -> CHOOSE(x'', cons(v'', w''), s(y''), s(z''))
INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(s(y'''''')), cons(v, cons(s(s(z'''''')), w'''''')), 0, s(z)) -> INSERT(s(s(y'''''')), cons(s(s(z'''''')), w''''''))
CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))
INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

CHOOSE(x'', cons(v'', w''), s(s(y'')), s(s(z''))) -> CHOOSE(x'', cons(v'', w''), s(y''), s(z''))

two new Dependency Pairs are created:

CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(s(0)), s(s(s(z'''')))) -> CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(s(0)), s(s(s(z'''')))) -> CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))
CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))
INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(s(y'''''')), cons(v, cons(s(s(z'''''')), w'''''')), 0, s(z)) -> INSERT(s(s(y'''''')), cons(s(s(z'''''')), w''''''))
CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))
INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))
CHOOSE(s(0), cons(v, cons(s(s(z'''''')), cons(s(z''''''''''), w''''''''''))), 0, s(z)) -> INSERT(s(0), cons(s(s(z'''''')), cons(s(z''''''''''), w'''''''''')))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), s(0), s(s(z''))) -> CHOOSE(s(y''''''), cons(v'', cons(s(z''''''), w'''''')), 0, s(z''))

two new Dependency Pairs are created:

CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), s(0), s(s(z'''))) -> CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), 0, s(z'''))
CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), s(0), s(s(z'''))) -> CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), 0, s(z'''))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 12

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))
CHOOSE(s(0), cons(v, cons(s(s(z'''''')), cons(s(z''''''''''), w''''''''''))), 0, s(z)) -> INSERT(s(0), cons(s(s(z'''''')), cons(s(z''''''''''), w'''''''''')))
CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), s(0), s(s(z'''))) -> CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), 0, s(z'''))

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

CHOOSE(s(0), cons(v, cons(s(s(z'''''')), cons(s(z''''''''''), w''''''''''))), 0, s(z)) -> INSERT(s(0), cons(s(s(z'''''')), cons(s(z''''''''''), w'''''''''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(CHOOSE(x₁, x₂, x₃, x₄)) = x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(INSERT(x₁, x₂)) = x₂

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 15

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

INSERT(s(0), cons(s(s(z'''')), cons(s(z''''''''), w''''''''))) -> CHOOSE(s(0), cons(s(s(z'''')), cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))
CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), s(0), s(s(z'''))) -> CHOOSE(s(0), cons(v''', cons(s(s(z'''''''')), cons(s(z''''''''''''), w''''''''''''))), 0, s(z'''))

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 13

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))
INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(s(y'''''')), cons(v, cons(s(s(z'''''')), w'''''')), 0, s(z)) -> INSERT(s(s(y'''''')), cons(s(s(z'''''')), w''''''))
CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), s(0), s(s(z'''))) -> CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), 0, s(z'''))
CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(s(0)), s(s(s(z'''')))) -> CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

CHOOSE(s(s(y'''''')), cons(v, cons(s(s(z'''''')), w'''''')), 0, s(z)) -> INSERT(s(s(y'''''')), cons(s(s(z'''''')), w''''''))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(CHOOSE(x₁, x₂, x₃, x₄)) = x₂

POL(0) = 0

POL(cons(x₁, x₂)) = 1 + x₂

POL(INSERT(x₁, x₂)) = x₂

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 16

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pairs:

CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))
INSERT(s(s(y'''')), cons(s(s(z'''')), w'''')) -> CHOOSE(s(s(y'''')), cons(s(s(z'''')), w''''), s(s(y'''')), s(s(z'''')))
CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), s(0), s(s(z'''))) -> CHOOSE(s(s(y'''''''')), cons(v''', cons(s(s(z'''''''')), w'''''''')), 0, s(z'''))
CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(s(0)), s(s(s(z'''')))) -> CHOOSE(s(y''''''''), cons(v'''', cons(s(z''''''''), w'''''''')), s(0), s(s(z'''')))

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)

Strategy:

innermost

Using the Dependency Graph the DP problem was split into 1 DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 17

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

CHOOSE(x'''', cons(v'''', w''''), s(s(s(y''''))), s(s(s(z'''')))) -> CHOOSE(x'''', cons(v'''', w''''), s(s(y'''')), s(s(z'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(CHOOSE(x₁, x₂, x₃, x₄)) = 1 + x₁ + x₃

POL(cons(x₁, x₂)) = 0

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 18

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pair:

SORT(cons(x, y)) -> SORT(y)

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(x, y)) -> SORT(y)

one new Dependency Pair is created:

SORT(cons(x, cons(x'', y''))) -> SORT(cons(x'', y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 19

             ↳Forward Instantiation Transformation

Dependency Pair:

SORT(cons(x, cons(x'', y''))) -> SORT(cons(x'', y''))

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)

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

SORT(cons(x, cons(x'', y''))) -> SORT(cons(x'', y''))

one new Dependency Pair is created:

SORT(cons(x, cons(x'''', cons(x''''', y'''')))) -> SORT(cons(x'''', cons(x''''', y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 19

             ↳FwdInst

...

               →DP Problem 20

                 ↳Polynomial Ordering

Dependency Pair:

SORT(cons(x, cons(x'''', cons(x''''', y'''')))) -> SORT(cons(x'''', cons(x''''', y'''')))

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)

Strategy:

innermost

The following dependency pair can be strictly oriented:

SORT(cons(x, cons(x'''', cons(x''''', y'''')))) -> SORT(cons(x'''', cons(x''''', y'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(cons(x₁, x₂)) = 1 + x₂

POL(SORT(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 19

             ↳FwdInst

...

               →DP Problem 21

                 ↳Dependency Graph

Dependency Pair:

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)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(CHOOSE(x₁, x₂, x₃, x₄))	= x₂
POL(0)	= 0
POL(cons(x₁, x₂))	= x₁ + x₂
POL(INSERT(x₁, x₂))	= x₂
POL(s(x₁))	= 1

POL(CHOOSE(x₁, x₂, x₃, x₄))	= 1 + x₁ + x₃
POL(cons(x₁, x₂))	= 0
POL(s(x₁))	= 1 + x₁

POL(cons(x₁, x₂))	= 1 + x₂
POL(SORT(x₁))	= 1 + x₁