Term Rewriting System R:
[y, x, f, xs]
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) -> nil
app(app(sumwith, f), app(app(cons, x), xs)) -> app(app(plus, app(f, x)), app(app(sumwith, f), xs))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(app(plus, app(f, x)), app(app(sumwith, f), xs))
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(plus, app(f, x))
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(app(sumwith, f), xs)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) -> nil
app(app(sumwith, f), app(app(cons, x), xs)) -> app(app(plus, app(f, x)), app(app(sumwith, f), xs))


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳A-Transformation
       →DP Problem 2
         ↳UsableRules


Dependency Pair:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rule:

none


Strategy:

innermost




We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 3
             ↳ATrans
             ...
               →DP Problem 4
                 ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules


Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)


Rule:

none


Strategy:

innermost




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

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

We obtain no new DP problems. 


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


Dependency Pairs:

APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(app(sumwith, f), xs)
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(f, x)


Rules:


app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(sumwith, f), nil) -> nil
app(app(sumwith, f), app(app(cons, x), xs)) -> app(app(plus, app(f, x)), app(app(sumwith, f), xs))


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 5
             ↳Size-Change Principle


Dependency Pairs:

APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(app(sumwith, f), xs)
APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(f, x)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(app(sumwith, f), xs)
  2. APP(app(sumwith, f), app(app(cons, x), xs)) -> APP(f, x)
and get the following Size-Change Graph(s):
{1, 2} , {1, 2}
1=1
2>2
{1, 2} , {1, 2}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1, 2} , {1, 2}
1=1
2>2
{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:
app(x1, x2) -> app(x1, x2)

We obtain no new DP problems.

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