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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

REV(cons(x, l)) -> REV1(x, l)
REV(cons(x, l)) -> REV2(x, l)
REV1(x, cons(y, l)) -> REV1(y, l)
REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))
REV2(x, cons(y, l)) -> REV2(y, l)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 2`
`         ↳UsableRules`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 3`
`             ↳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:
1. 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

DP: 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.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳Usable Rules (Innermost)`

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:

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

Strategy:

innermost

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

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳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:

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

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

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( s(x1) ) = 0

POL( rev(x1) ) = x1

POL( nil ) = 0

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

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

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes