Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

BSORT(.(x, y)) -> LAST(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
BSORT(.(x, y)) -> BUBBLE(.(x, y))
BSORT(.(x, y)) -> BSORT(butlast(bubble(.(x, y))))
BSORT(.(x, y)) -> BUTLAST(bubble(.(x, y)))
BUBBLE(.(x, .(y, z))) -> BUBBLE(.(x, z))
BUBBLE(.(x, .(y, z))) -> BUBBLE(.(y, z))
LAST(.(x, .(y, z))) -> LAST(.(y, z))
BUTLAST(.(x, .(y, z))) -> BUTLAST(.(y, z))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

LAST(.(x, .(y, z))) -> LAST(.(y, z))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

LAST(.(x, .(y, z))) -> LAST(.(y, z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LAST(x₁)) = 1 + x₁

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polynomial Ordering

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pairs:

BUBBLE(.(x, .(y, z))) -> BUBBLE(.(y, z))
BUBBLE(.(x, .(y, z))) -> BUBBLE(.(x, z))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

BUBBLE(.(x, .(y, z))) -> BUBBLE(.(y, z))
BUBBLE(.(x, .(y, z))) -> BUBBLE(.(x, z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(BUBBLE(x₁)) = 1 + x₁

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polynomial Ordering

       →DP Problem 4

         ↳Nar

Dependency Pair:

BUTLAST(.(x, .(y, z))) -> BUTLAST(.(y, z))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

BUTLAST(.(x, .(y, z))) -> BUTLAST(.(y, z))

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(BUTLAST(x₁)) = 1 + x₁

POL(.(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Narrowing Transformation

Dependency Pair:

BSORT(.(x, y)) -> BSORT(butlast(bubble(.(x, y))))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

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

BSORT(.(x, y)) -> BSORT(butlast(bubble(.(x, y))))

two new Dependency Pairs are created:

BSORT(.(x'', nil)) -> BSORT(butlast(.(x'', nil)))
BSORT(.(x'', .(y'', z'))) -> BSORT(butlast(if(<=(x'', y''), .(y'', bubble(.(x'', z'))), .(x'', bubble(.(y'', z'))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rewriting Transformation

Dependency Pairs:

BSORT(.(x'', .(y'', z'))) -> BSORT(butlast(if(<=(x'', y''), .(y'', bubble(.(x'', z'))), .(x'', bubble(.(y'', z'))))))
BSORT(.(x'', nil)) -> BSORT(butlast(.(x'', nil)))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

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

BSORT(.(x'', nil)) -> BSORT(butlast(.(x'', nil)))

one new Dependency Pair is created:

BSORT(.(x'', nil)) -> BSORT(nil)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 9

                 ↳Polynomial Ordering

Dependency Pair:

BSORT(.(x'', .(y'', z'))) -> BSORT(butlast(if(<=(x'', y''), .(y'', bubble(.(x'', z'))), .(x'', bubble(.(y'', z'))))))

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

BSORT(.(x'', .(y'', z'))) -> BSORT(butlast(if(<=(x'', y''), .(y'', bubble(.(x'', z'))), .(x'', bubble(.(y'', z'))))))

Additionally, the following usable rules for innermost can be oriented:

butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(if(x₁, x₂, x₃)) = 0

POL(butlast(x₁)) = x₁

POL(nil) = 0

POL(.(x₁, x₂)) = 1

POL(bubble(x₁)) = x₁

POL(BSORT(x₁)) = 1 + x₁

POL(<=(x₁, x₂)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Polo

       →DP Problem 3

         ↳Polo

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Rw

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pair:

Rules:

bsort(nil) -> nil
bsort(.(x, y)) -> last(.(bubble(.(x, y)), bsort(butlast(bubble(.(x, y))))))
bubble(nil) -> nil
bubble(.(x, nil)) -> .(x, nil)
bubble(.(x, .(y, z))) -> if(<=(x, y), .(y, bubble(.(x, z))), .(x, bubble(.(y, z))))
last(nil) -> 0
last(.(x, nil)) -> x
last(.(x, .(y, z))) -> last(.(y, z))
butlast(nil) -> nil
butlast(.(x, nil)) -> nil
butlast(.(x, .(y, z))) -> .(x, butlast(.(y, z)))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(LAST(x₁))	= 1 + x₁
POL(.(x₁, x₂))	= 1 + x₂

POL(BUBBLE(x₁))	= 1 + x₁
POL(.(x₁, x₂))	= 1 + x₂

POL(BUTLAST(x₁))	= 1 + x₁
POL(.(x₁, x₂))	= 1 + x₂

POL(if(x₁, x₂, x₃))	= 0
POL(butlast(x₁))	= x₁
POL(nil)	= 0
POL(.(x₁, x₂))	= 1
POL(bubble(x₁))	= x₁
POL(BSORT(x₁))	= 1 + x₁
POL(<=(x₁, x₂))	= 0