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
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:



   R
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:



   R
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 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.



   R
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.


   R
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:



   R
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:



   R
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:



   R
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:



   R
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:



   R
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:



   R
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 w.r.t. to the implicit AFS can be oriented:

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(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.



   R
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