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.


This program has no overlaps, so it is sufficient to show innermost termination. 


R contains the following Dependency Pairs: 


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

Furthermore, R contains two SCCs.


SCC1:

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

Removing rules from R by ordering and analyzing Dependency Pairs, Usable Rules, and Usable Equations.

This is possible by using the following (C_E-compatible) Polynomial ordering.

Polynomial interpretation:
POL(REV1(x_1, x_2)) = 1 + x_1 + x_2
POL(cons(x_1, x_2)) = 1 + x_1 + x_2

The following Dependency Pairs can be deleted:

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



 This transformation is resulting in no new subcycles.


SCC2:

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

By using a polynomial ordering, at least one Dependency Pair of this SCC can be strictly oriented.

Additionally, the following rules can be oriented: 

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


Used ordering: Polynomial ordering with Polynomial interpretation:
POL(nil) = 0
POL(s(x_1)) = 0
POL(rev2(x_1, x_2)) = x_2
POL(REV(x_1)) = x_1
POL(rev1(x_1, x_2)) = 0
POL(rev(x_1)) = x_1
POL(REV2(x_1, x_2)) = x_2
POL(0) = 0
POL(cons(x_1, x_2)) = 1 + x_2

 resulting in no subcycles.




Termination of R successfully shown.


Duration: 0.764 seconds.