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
     ↳Overlay and local confluence Check



The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost. 


   R
     ↳OC
       →TRS2
         ↳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
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳Usable Rules (Innermost)
           →DP Problem 2
             ↳UsableRules


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




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
             ...
               →DP Problem 3
                 ↳Size-Change Principle
           →DP Problem 2
             ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. REV1(x, cons(y, l)) -> REV1(y, l)
and get the following Size-Change Graph(s):
{1} , {1}
2>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
2>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳Usable Rules (Innermost)


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




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 4
                 ↳Negative Polynomial Order


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:


rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))


Strategy:

innermost




The following Dependency Pairs can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))


Used ordering:
Polynomial Order with Interpretation:

POL( REV2(x1, x2) ) = x2

POL( cons(x1, x2) ) = x2 + 1

POL( REV(x1) ) = x1

POL( rev2(x1, x2) ) = x2

POL( rev1(x1, x2) ) = 0

POL( 0 ) = 0

POL( s(x1) ) = 0

POL( rev(x1) ) = x1

POL( nil ) = 0


This results in one new DP problem. 


   R
     ↳OC
       →TRS2
         ↳DPs
           →DP Problem 1
             ↳UsableRules
           →DP Problem 2
             ↳UsableRules
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:

REV2(x, cons(y, l)) -> REV(cons(x, rev2(y, l)))


Rules:


rev1(0, nil) -> 0
rev1(s(x), nil) -> s(x)
rev1(x, cons(y, l)) -> rev1(y, l)
rev2(x, cons(y, l)) -> rev(cons(x, rev2(y, l)))
rev2(x, nil) -> nil
rev(cons(x, l)) -> cons(rev1(x, l), rev2(x, l))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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