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)))

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
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


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)))





The following dependency pair can be strictly oriented:

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


The following rules can be oriented:

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)))


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
nil > 0
nil > s

resulting in one new DP problem.
Used Argument Filtering System:
REV1(x1, x2) -> x2
cons(x1, x2) -> cons(x2)
rev(x1) -> x1
rev2(x1, x2) -> x2
rev1(x1, x2) -> x2
s(x1) -> s


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


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)))





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(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)))





The following dependency pairs can be strictly oriented:

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


The following rules 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))
rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{nil, s} > 0

resulting in one new DP problem.
Used Argument Filtering System:
REV2(x1, x2) -> x2
cons(x1, x2) -> cons(x2)
REV(x1) -> x1
rev2(x1, x2) -> x2
rev(x1) -> x1
rev1(x1, x2) -> x2
s(x1) -> s


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳Dependency Graph


Dependency Pair:

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





Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:02 minutes