Termination Proof using AProVE

Term Rewriting System R:
[x, l, y]
app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(rev, app(app(cons, x), l)) -> APP(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
APP(rev, app(app(cons, x), l)) -> APP(cons, app(app(rev1, x), l))
APP(rev, app(app(cons, x), l)) -> APP(app(rev1, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev1, x)
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)
APP(rev, app(app(cons, x), l)) -> APP(rev2, x)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)
APP(app(rev1, x), app(app(cons, y), l)) -> APP(rev1, y)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(cons, x), app(app(rev2, y), l))
APP(app(rev2, x), app(app(cons, y), l)) -> APP(cons, x)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev2, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)

Rules:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Strategy:

innermost

As we are in the innermost case, we can delete all 7 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳A-Transformation

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

APP(app(rev1, x), app(app(cons, y), l)) -> APP(app(rev1, y), l)

Rule:

none

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳ATrans

...

               →DP Problem 4

                 ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

Dependency Pair:

REV1(x, cons(y, l)) -> REV1(y, l)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

REV1(x, cons(y, l)) -> REV1(y, l)

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:

cons(x₁, x₂) -> cons(x₁, x₂)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

Dependency Pairs:

APP(app(rev2, x), app(app(cons, y), l)) -> APP(app(rev2, y), l)
APP(app(rev2, x), app(app(cons, y), l)) -> APP(rev, app(app(cons, x), app(app(rev2, y), l)))
APP(rev, app(app(cons, x), l)) -> APP(app(rev2, x), l)

Rules:

app(rev, nil) -> nil
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, 0), nil) -> 0
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(app(rev2, x), nil) -> nil
app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))

Strategy:

innermost

As we are in the innermost case, we can delete all 1 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳A-Transformation

Dependency Pairs:

Rules:

app(app(rev2, x), app(app(cons, y), l)) -> app(rev, app(app(cons, x), app(app(rev2, y), l)))
app(app(rev1, 0), nil) -> 0
app(app(rev1, x), app(app(cons, y), l)) -> app(app(rev1, y), l)
app(rev, app(app(cons, x), l)) -> app(app(cons, app(app(rev1, x), l)), app(app(rev2, x), l))
app(app(rev1, app(s, x)), nil) -> app(s, x)
app(app(rev2, x), nil) -> nil

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 6

                 ↳Negative Polynomial Order

Dependency Pairs:

REV2(x, cons(y, l)) -> REV2(y, l)
REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
REV(cons(x, l)) -> REV2(x, l)

Rules:

rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(x, cons(y, l)) -> rev1(y, l)
rev1(s(x), nil) -> s(x)

Strategy:

innermost

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

REV2(x, cons(y, l)) -> REV2(y, l)
REV(cons(x, l)) -> REV2(x, l)

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

rev1(0, nil) -> 0
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev1(s(x), nil) -> s(x)
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(x, cons(y, l)) -> rev1(y, l)

Used ordering:
Polynomial Order with Interpretation:

POL( REV2(x₁, x₂) ) = x₂

POL( cons(x₁, x₂) ) = x₂ + 1

POL( REV(x₁) ) = x₁

POL( rev2(x₁, x₂) ) = x₂

POL( rev1(x₁, x₂) ) = 0

POL( 0 ) = 0

POL( rev(x₁) ) = x₁

POL( nil ) = 0

POL( s(x₁) ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳ATrans

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pair:

REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))

Rules:

rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))
rev1(0, nil) -> 0
rev1(x, cons(y, l)) -> rev1(y, l)
rev1(s(x), nil) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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