Term Rewriting System R:
[n, x, m, y]
sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n

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:

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
SUM(cons(0, x), y) -> SUM(x, y)
WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))
WEIGHT(cons(n, cons(m, x))) -> SUM(cons(n, cons(m, x)), cons(0, x))

Furthermore, R contains two SCCs. 


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


Dependency Pairs:

SUM(cons(0, x), y) -> SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))


Rules:


sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n


Strategy:

innermost




As we are in the innermost case, we can delete all 5 non-usable-rules.


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


Dependency Pairs:

SUM(cons(0, x), y) -> SUM(x, y)
SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. SUM(cons(0, x), y) -> SUM(x, y)
  2. SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1>1
2=2
{1, 2} , {1, 2}
1>1

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1>1
{1, 2} , {1, 2}
1>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)
s(x1) -> s(x1)

We obtain no new DP problems. 


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


Dependency Pair:

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))


Rules:


sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n


Strategy:

innermost




As we are in the innermost case, we can delete all 2 non-usable-rules.


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


Dependency Pair:

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))


Rules:


sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y


Strategy:

innermost




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

WEIGHT(cons(n, cons(m, x))) -> WEIGHT(sum(cons(n, cons(m, x)), cons(0, x)))


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

sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y


Used ordering:
Polynomial Order with Interpretation:

POL( WEIGHT(x1) ) = x1

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

POL( sum(x1, x2) ) = x2


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:


Rules:


sum(cons(s(n), x), cons(m, y)) -> sum(cons(n, x), cons(s(m), y))
sum(cons(0, x), y) -> sum(x, y)
sum(nil, y) -> y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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