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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


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 AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
cons > REV1

resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> REV1(x1, x2)
cons(x1, x2) -> cons(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS


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
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering


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))))





The following dependency pairs can be strictly oriented:

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)


The following usable rules w.r.t. to the 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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
cons > REV2 > REV

resulting in one new DP problem.
Used Argument Filtering System:
REV(x1) -> REV(x1)
REV2(x1, x2) -> REV2(x2)
cons(x1, x2) -> cons(x2)
rev(x1) -> x1
rev2(x1, x2) -> x2


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳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))))





Using the Dependency Graph resulted in no new DP problems.

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