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)))

Termination of R to be shown.

     ↳Overlay and local confluence Check

The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.

     ↳OC

       →TRS2

         ↳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.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Usable Rules (Innermost)

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

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

As we are in the innermost case, we can delete all 11 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

...

               →DP Problem 5

                 ↳Size-Change Principle

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

.(x₁, x₂) -> .(x₁, x₂)

We obtain no new DP problems.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳Usable Rules (Innermost)

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

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

As we are in the innermost case, we can delete all 11 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

...

               →DP Problem 6

                 ↳Size-Change Principle

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1, 2}	,	{1, 2}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1, 2}	,	{1, 2}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

.(x₁, x₂) -> .(x₁, x₂)

We obtain no new DP problems.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳Usable Rules (Innermost)

           →DP Problem 4

             ↳UsableRules

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

As we are in the innermost case, we can delete all 11 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 7

                 ↳Size-Change Principle

           →DP Problem 4

             ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

.(x₁, x₂) -> .(x₁, x₂)

We obtain no new DP problems.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳Usable Rules (Innermost)

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

As we are in the innermost case, we can delete all 6 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

           →DP Problem 4

             ↳UsableRules

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pair:

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

Rules:

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

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

The transformation is resulting in no new DP problems.

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