Termination Proof using AProVE

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.

     ↳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.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

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

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

         ↳FwdInst

Dependency Pair:

REV1(x, cons(y0, cons(y'', l''))) -> REV1(y0, cons(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

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

         ↳FwdInst

Dependency Pair:

REV1(x, cons(y0'', cons(y''0, cons(y'''', l'''')))) -> REV1(y0'', cons(y''0, cons(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

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 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

         ↳FwdInst

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

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

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)

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

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

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, rev2(y, l)))

two new Dependency Pairs are created:

REV2(x, cons(y', nil)) -> REV(cons(x, nil))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, rev(cons(y0, rev2(y'', l'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Rewriting Transformation

Dependency Pairs:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, rev(cons(y0, rev2(y'', l'')))))
REV2(x, cons(y, l)) -> REV2(y, l)
REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(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

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(y0, rev2(y'', l'')))))

one new Dependency Pair is created:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

Dependency Pairs:

REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(y'', l''))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV2(x, cons(y, l)) -> REV2(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

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

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳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''))
REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV(cons(x'', cons(y'', l''))) -> REV2(x'', cons(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

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

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳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(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV2(x, cons(y', cons(y0'', cons(y'''', l'''')))) -> REV2(y', cons(y0'', cons(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

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

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳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(y0, rev2(y'', l'')), rev2(y0, rev2(y'', l'')))))
REV(cons(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(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

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(y'''', cons(y0'''', cons(y'''''', l''''''))))) -> REV2(x''', cons(y'''', cons(y0'''', cons(y'''''', l''''''))))

Additionally, the following usable rules for innermost can be oriented:

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)))
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(0) = 0

POL(rev(x₁)) = x₁

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

         ↳FwdInst

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

REV2(x, cons(y0, cons(y'', l''))) -> REV(cons(x, cons(rev1(y0, rev2(y'', l'')), rev2(y0, rev2(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

Using the Dependency Graph resulted in no new DP problems.

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

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

POL(rev2(x₁, x₂))	= x₂
POL(0)	= 0
POL(rev(x₁))	= x₁
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