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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(sum, app(app(cons, x), xs)) -> APP(app(plus, x), app(sum, xs))
APP(sum, app(app(cons, x), xs)) -> APP(plus, x)
APP(sum, app(app(cons, x), xs)) -> APP(sum, xs)
APP(size, app(app(node, x), xs)) -> APP(s, app(sum, app(app(map, size), xs)))
APP(size, app(app(node, x), xs)) -> APP(sum, app(app(map, size), xs))
APP(size, app(app(node, x), xs)) -> APP(app(map, size), xs)
APP(size, app(app(node, x), xs)) -> APP(map, size)
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)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(size, app(app(node, x), xs)) -> APP(app(map, size), xs)
APP(size, app(app(node, x), xs)) -> APP(sum, app(app(map, size), xs))
APP(sum, app(app(cons, x), xs)) -> APP(sum, xs)
APP(sum, app(app(cons, x), xs)) -> APP(app(plus, x), app(sum, xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))


Rules:


app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(sum, app(app(cons, x), xs)) -> app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) -> app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) -> 0
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))


Strategy:

innermost




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

APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
eight new Dependency Pairs are created:

APP(app(map, app(map, f'')), app(app(cons, nil), xs)) -> APP(app(cons, nil), app(app(map, app(map, f'')), xs))
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, sum), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(plus, x''), app(sum, xs''))), app(app(map, sum), xs))
APP(app(map, size), app(app(cons, app(app(node, x''), xs'')), xs)) -> APP(app(cons, app(s, app(sum, app(app(map, size), xs'')))), app(app(map, size), xs))
APP(app(map, app(plus, 0)), app(app(cons, x''), xs)) -> APP(app(cons, 0), app(app(map, app(plus, 0)), xs))
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Narrowing Transformation


Dependency Pairs:

APP(app(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))
APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, app(plus, 0)), app(app(cons, x''), xs)) -> APP(app(cons, 0), app(app(map, app(plus, 0)), xs))
APP(app(map, size), app(app(cons, app(app(node, x''), xs'')), xs)) -> APP(app(cons, app(s, app(sum, app(app(map, size), xs'')))), app(app(map, size), xs))
APP(app(map, sum), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(plus, x''), app(sum, xs''))), app(app(map, sum), xs))
APP(size, app(app(node, x), xs)) -> APP(app(map, size), xs)
APP(size, app(app(node, x), xs)) -> APP(sum, app(app(map, size), xs))
APP(sum, app(app(cons, x), xs)) -> APP(sum, xs)
APP(sum, app(app(cons, x), xs)) -> APP(app(plus, x), app(sum, xs))
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)


Rules:


app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(sum, app(app(cons, x), xs)) -> app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) -> app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) -> 0
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))


Strategy:

innermost




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

APP(size, app(app(node, x), xs)) -> APP(sum, app(app(map, size), xs))
two new Dependency Pairs are created:

APP(size, app(app(node, x), nil)) -> APP(sum, nil)
APP(size, app(app(node, x), app(app(cons, x''), xs''))) -> APP(sum, app(app(cons, app(size, x'')), app(app(map, size), xs'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(size, app(app(node, x), app(app(cons, x''), xs''))) -> APP(sum, app(app(cons, app(size, x'')), app(app(map, size), xs'')))
APP(app(map, f''), app(app(cons, x), nil)) -> APP(app(cons, app(f'', x)), nil)
APP(app(map, app(plus, app(s, x''))), app(app(cons, x0), xs)) -> APP(app(cons, app(s, app(app(plus, x''), x0))), app(app(map, app(plus, app(s, x''))), xs))
APP(app(map, app(plus, 0)), app(app(cons, x''), xs)) -> APP(app(cons, 0), app(app(map, app(plus, 0)), xs))
APP(app(map, size), app(app(cons, app(app(node, x''), xs'')), xs)) -> APP(app(cons, app(s, app(sum, app(app(map, size), xs'')))), app(app(map, size), xs))
APP(app(map, sum), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(plus, x''), app(sum, xs''))), app(app(map, sum), xs))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(size, app(app(node, x), xs)) -> APP(app(map, size), xs)
APP(sum, app(app(cons, x), xs)) -> APP(sum, xs)
APP(sum, app(app(cons, x), xs)) -> APP(app(plus, x), app(sum, xs))
APP(app(map, app(map, f'')), app(app(cons, app(app(cons, x''), xs'')), xs)) -> APP(app(cons, app(app(cons, app(f'', x'')), app(app(map, f''), xs''))), app(app(map, app(map, f'')), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f''), app(app(cons, x), app(app(cons, x''), xs''))) -> APP(app(cons, app(f'', x)), app(app(cons, app(f'', x'')), app(app(map, f''), xs'')))


Rules:


app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
app(sum, app(app(cons, x), xs)) -> app(app(plus, x), app(sum, xs))
app(size, app(app(node, x), xs)) -> app(s, app(sum, app(app(map, size), xs)))
app(app(plus, 0), x) -> 0
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:15 minutes