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)

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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

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)

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

CHOOSE(x₁, x₂, x₃, x₄) -> x₂
cons(x₁, x₂) -> cons(x₁, x₂)
INSERT(x₁, x₂) -> x₂

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

Dependency Pairs:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 4

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

Dependency Pair:

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)

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

CHOOSE(x₁, x₂, x₃, x₄) -> CHOOSE(x₁, x₂, x₃, x₄)
cons(x₁, x₂) -> cons(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳DGraph

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳AFS

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)

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

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)

The following dependency pair can be strictly oriented:

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

There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

SORT(x₁) -> SORT(x₁)
cons(x₁, x₂) -> cons(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳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)

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:00 minutes