Termination Proof using AProVE

Term Rewriting System R:
[X, Y, L]
rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

REV1(X, cons(Y, L)) -> REV1(Y, L)
REV(cons(X, L)) -> REV1(X, L)
REV(cons(X, L)) -> REV2(X, L)
REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) -> REV(rev2(Y, L))
REV2(X, cons(Y, L)) -> REV2(Y, L)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

REV1(X, cons(Y, L)) -> REV1(Y, L)

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV1(X, cons(Y, L)) -> REV1(Y, L)

one new Dependency Pair is created:

REV1(X, cons(Y0, cons(Y'', L''))) -> REV1(Y0, cons(Y'', L''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

REV1(X, cons(Y0, cons(Y'', L''))) -> REV1(Y0, cons(Y'', L''))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV1(X, cons(Y0, cons(Y'', L''))) -> REV1(Y0, cons(Y'', L''))

one new Dependency Pair is created:

REV1(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV1(Y0'', cons(Y''0, cons(Y'''', L'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

REV1(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV1(Y0'', cons(Y''0, cons(Y'''', L'''')))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

REV1(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV1(Y0'', cons(Y''0, cons(Y'''', L'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(REV1(x₁, x₂)) = x₂

POL(cons(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

REV2(X, cons(Y, L)) -> REV2(Y, L)
REV2(X, cons(Y, L)) -> REV(rev2(Y, L))
REV2(X, cons(Y, L)) -> REV(cons(X, rev(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)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y, L)) -> REV(rev2(Y, L))

two new Dependency Pairs are created:

REV2(X, cons(Y', nil)) -> REV(nil)
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(rev(cons(Y0, rev(rev2(Y'', L'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rewriting Transformation

Dependency Pairs:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(rev(cons(Y0, rev(rev2(Y'', L'')))))
REV(cons(X, L)) -> REV2(X, L)
REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L))))
REV2(X, cons(Y, L)) -> REV2(Y, L)

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(rev(cons(Y0, rev(rev2(Y'', L'')))))

one new Dependency Pair is created:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pairs:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y, L)) -> REV2(Y, L)
REV2(X, cons(Y, L)) -> REV(cons(X, rev(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)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV(cons(X, L)) -> REV2(X, L)

two new Dependency Pairs are created:

REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

REV2(X, cons(Y, L)) -> REV2(Y, L)
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L))))
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y, L)) -> REV(cons(X, rev(rev2(Y, L))))

two new Dependency Pairs are created:

REV2(X, cons(Y', nil)) -> REV(cons(X, rev(nil)))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(rev(cons(Y0, rev(rev2(Y'', L'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 9

                 ↳Rewriting Transformation

Dependency Pairs:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(rev(cons(Y0, rev(rev2(Y'', L'')))))))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y', nil)) -> REV(cons(X, rev(nil)))
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y, L)) -> REV2(Y, L)

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y', nil)) -> REV(cons(X, rev(nil)))

one new Dependency Pair is created:

REV2(X, cons(Y', nil)) -> REV(cons(X, nil))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pairs:

REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y, L)) -> REV2(Y, L)
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(rev(cons(Y0, rev(rev2(Y'', L'')))))))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(rev(cons(Y0, rev(rev2(Y'', L'')))))))

one new Dependency Pair is created:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 11

                 ↳Rewriting Transformation

Dependency Pairs:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y, L)) -> REV2(Y, L)
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, rev(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))

one new Dependency Pair is created:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, cons(rev1(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))), rev2(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 12

                 ↳Forward Instantiation Transformation

Dependency Pairs:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, cons(rev1(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))), rev2(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y, L)) -> REV2(Y, L)
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y, L)) -> REV2(Y, L)

two new Dependency Pairs are created:

REV2(X, cons(Y0, cons(Y'', L''))) -> REV2(Y0, cons(Y'', L''))
REV2(X, cons(Y', cons(Y0'', cons(Y'''', L'''')))) -> REV2(Y', cons(Y0'', cons(Y'''', L'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

Dependency Pairs:

REV2(X, cons(Y', cons(Y0'', cons(Y'''', L'''')))) -> REV2(Y', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV2(Y0, cons(Y'', L''))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, cons(rev1(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))), rev2(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV(cons(X'', cons(Y'', L''))) -> REV2(X'', cons(Y'', L''))

two new Dependency Pairs are created:

REV(cons(X''', cons(Y''0, cons(Y'''', L'''')))) -> REV2(X''', cons(Y''0, cons(Y'''', L'''')))
REV(cons(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 14

                 ↳Forward Instantiation Transformation

Dependency Pairs:

REV(cons(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV2(Y0, cons(Y'', L''))
REV(cons(X''', cons(Y''0, cons(Y'''', L'''')))) -> REV2(X''', cons(Y''0, cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, cons(rev1(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))), rev2(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV2(X, cons(Y', cons(Y0'', cons(Y'''', L'''')))) -> REV2(Y', cons(Y0'', cons(Y'''', L'''')))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

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

REV2(X, cons(Y0, cons(Y'', L''))) -> REV2(Y0, cons(Y'', L''))

two new Dependency Pairs are created:

REV2(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV2(Y0'', cons(Y''0, cons(Y'''', L'''')))
REV2(X, cons(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 15

                 ↳Polynomial Ordering

Dependency Pairs:

REV2(X, cons(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))
REV2(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV2(Y0'', cons(Y''0, cons(Y'''', L'''')))
REV2(X, cons(Y', cons(Y0'', cons(Y'''', L'''')))) -> REV2(Y', cons(Y0'', cons(Y'''', L'''')))
REV(cons(X''', cons(Y''0, cons(Y'''', L'''')))) -> REV2(X''', cons(Y''0, cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(X, cons(rev1(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))), rev2(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV(cons(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

REV2(X, cons(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(Y0', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))
REV2(X, cons(Y0'', cons(Y''0, cons(Y'''', L'''')))) -> REV2(Y0'', cons(Y''0, cons(Y'''', L'''')))
REV2(X, cons(Y', cons(Y0'', cons(Y'''', L'''')))) -> REV2(Y', cons(Y0'', cons(Y'''', L'''')))
REV(cons(X''', cons(Y''0, cons(Y'''', L'''')))) -> REV2(X''', cons(Y''0, cons(Y'''', L'''')))
REV(cons(X'', cons(Y0'', cons(Y'''', L'''')))) -> REV2(X'', cons(Y0'', cons(Y'''', L'''')))
REV2(X, cons(Y0, cons(Y'', L''))) -> REV(cons(rev1(Y0, rev(rev2(Y'', L''))), rev2(Y0, rev(rev2(Y'', L'')))))
REV(cons(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))) -> REV2(X''', cons(Y'''', cons(Y0'''', cons(Y'''''', L''''''))))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(rev2(x₁, x₂)) = x₂

POL(rev(x₁)) = x₁

POL(0) = 0

POL(REV(x₁)) = x₁

POL(cons(x₁, x₂)) = 1 + x₂

POL(rev1(x₁, x₂)) = 0

POL(REV2(x₁, x₂)) = x₂

POL(nil) = 0

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Rw

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

Rules:

rev1(0, nil) -> 0
rev1(s(X), nil) -> s(X)
rev1(X, cons(Y, L)) -> rev1(Y, L)
rev(nil) -> nil
rev(cons(X, L)) -> cons(rev1(X, L), rev2(X, L))
rev2(X, nil) -> nil
rev2(X, cons(Y, L)) -> rev(cons(X, rev(rev2(Y, L))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(REV1(x₁, x₂))	= x₂
POL(cons(x₁, x₂))	= 1 + x₂

POL(rev2(x₁, x₂))	= x₂
POL(rev(x₁))	= x₁
POL(0)	= 0
POL(REV(x₁))	= x₁
POL(cons(x₁, x₂))	= 1 + x₂
POL(rev1(x₁, x₂))	= 0
POL(REV2(x₁, x₂))	= x₂
POL(nil)	= 0
POL(s(x₁))	= 0