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
     ↳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
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


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





The following dependency pair can be strictly oriented:

SUM(cons(0, x), y) -> SUM(x, y)


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SUM(x1, x2))=  x1  
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(s(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Instantiation Transformation
       →DP Problem 2
         ↳Polo


Dependency Pair:

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





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

SUM(cons(s(n), x), cons(m, y)) -> SUM(cons(n, x), cons(s(m), y))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 4
                 ↳Instantiation Transformation
       →DP Problem 2
         ↳Polo


Dependency Pair:

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





On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

SUM(cons(s(n''), x''), cons(s(m''), y'')) -> SUM(cons(n'', x''), cons(s(s(m'')), y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 5
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo


Dependency Pair:

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





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(SUM(x1, x2))=  1 + x1  
  POL(cons(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Inst
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 2
         ↳Polo


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering


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





The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules w.r.t. to the implicit AFS can be 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 ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(cons(x1, x2))=  1 + x2  
  POL(WEIGHT(x1))=  1 + x1  
  POL(nil)=  0  
  POL(sum(x1, x2))=  x2  
  POL(s(x1))=  0  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 7
             ↳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
weight(cons(n, cons(m, x))) -> weight(sum(cons(n, cons(m, x)), cons(0, x)))
weight(cons(n, nil)) -> n





Using the Dependency Graph resulted in no new DP problems.

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