Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(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:

REV(++(x, y)) -> REV1(x, y)
REV(++(x, y)) -> REV2(x, y)
REV1(x, ++(y, z)) -> REV1(y, z)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) -> REV(rev2(y, z))
REV2(x, ++(y, z)) -> REV2(y, z)

Furthermore, R contains two SCCs.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Usable Rules (Innermost)

           →DP Problem 2

             ↳Nar

Dependency Pair:

REV1(x, ++(y, z)) -> REV1(y, z)

Rules:

rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(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 3

                 ↳Size-Change Principle

           →DP Problem 2

             ↳Nar

Dependency Pair:

REV1(x, ++(y, z)) -> REV1(y, z)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

REV1(x, ++(y, z)) -> REV1(y, z)

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

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

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

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

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

             ↳Narrowing Transformation

Dependency Pairs:

REV2(x, ++(y, z)) -> REV2(y, z)
REV2(x, ++(y, z)) -> REV(rev2(y, z))
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV(++(x, y)) -> REV2(x, y)

Rules:

rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Strategy:

innermost

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

REV2(x, ++(y, z)) -> REV(rev2(y, z))

two new Dependency Pairs are created:

REV2(x, ++(y', nil)) -> REV(nil)
REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))

The transformation is resulting in one new DP problem:

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 4

                 ↳Negative Polynomial Order

Dependency Pairs:

REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))
REV(++(x, y)) -> REV2(x, y)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) -> REV2(y, z)

Rules:

rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

REV2(x, ++(y0, ++(y'', z''))) -> REV(rev(++(y0, rev(rev2(y'', z'')))))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))
rev(nil) -> nil
rev2(x, nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))

Used ordering:
Polynomial Order with Interpretation:

POL( REV2(x₁, x₂) ) = max{0, x₂ - 1}

POL( ++(x₁, x₂) ) = x₂ + 1

POL( REV(x₁) ) = max{0, x₁ - 1}

POL( rev(x₁) ) = x₁

POL( rev2(x₁, x₂) ) = x₂

POL( nil ) = 0

This results in one new DP problem.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 5

                 ↳Negative Polynomial Order

Dependency Pairs:

REV(++(x, y)) -> REV2(x, y)
REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))
REV2(x, ++(y, z)) -> REV2(y, z)

Rules:

rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Strategy:

innermost

The following Dependency Pairs can be strictly oriented using the given order.

REV(++(x, y)) -> REV2(x, y)
REV2(x, ++(y, z)) -> REV2(y, z)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))
rev(nil) -> nil
rev2(x, nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))

Used ordering:
Polynomial Order with Interpretation:

POL( REV(x₁) ) = x₁

POL( ++(x₁, x₂) ) = x₂ + 1

POL( REV2(x₁, x₂) ) = x₂

POL( rev(x₁) ) = x₁

POL( rev2(x₁, x₂) ) = x₂

POL( nil ) = 0

This results in one new DP problem.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳Nar

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

REV2(x, ++(y, z)) -> REV(++(x, rev(rev2(y, z))))

Rules:

rev(nil) -> nil
rev(++(x, y)) -> ++(rev1(x, y), rev2(x, y))
rev1(x, nil) -> x
rev1(x, ++(y, z)) -> rev1(y, z)
rev2(x, nil) -> nil
rev2(x, ++(y, z)) -> rev(++(x, rev(rev2(y, z))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:19 minutes