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))))

Termination of R to be shown.



   R
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.


   R
DPs
       →DP Problem 1
Polynomial Ordering
       →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))))





The following dependency pair can be strictly oriented:

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


There are no usable rules 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
Polo
           →DP Problem 3
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))))





Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
Polo
       →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))))





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:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
           →DP Problem 4
Narrowing 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))))





On this DP problem, a Narrowing 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'')))))
three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
           →DP Problem 4
Nar
             ...
               →DP Problem 5
Polynomial Ordering


Dependency Pairs:

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





The following dependency pairs can be strictly oriented:

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


Additionally, the following usable rules 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.



   R
DPs
       →DP Problem 1
Polo
       →DP Problem 2
Nar
           →DP Problem 4
Nar
             ...
               →DP Problem 6
Dependency Graph


Dependency Pair:

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





Using the Dependency Graph resulted in no new DP problems.

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