Term Rewriting System R:
[y, x, f, ys]
app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)
APP(app(neq, app(s, x)), app(s, y)) -> APP(neq, x)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(filtersub, app(f, y))
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(cons, y), app(app(filter, f), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(filter, f)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(filter, f)
NONZERO -> APP(filter, app(neq, 0))
NONZERO -> APP(neq, 0)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
eight new Dependency Pairs are created:

APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, f), app(app(cons, y), ys)) -> APP(app(filtersub, app(f, y)), f)
eight new Dependency Pairs are created:

APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(filtersub, false), app(neq, 0))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(filtersub, true), app(neq, 0))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(filtersub, true), app(neq, app(s, x')))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x')))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(filtersub, nil), app(filter, f''))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f''))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f''))
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 3
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, app(filter, f'')), app(app(cons, nil), ys)) -> APP(app(app(filtersub, nil), app(filter, f'')), app(app(cons, nil), ys))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Polynomial Ordering


Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

APP(app(neq, app(s, x)), app(s, y)) -> APP(app(neq, x), y)
APP(app(filter, f), app(app(cons, y), ys)) -> APP(f, y)


Additionally, the following usable rules for innermost can be oriented:

app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(filter)=  0  
  POL(0)=  0  
  POL(false)=  0  
  POL(cons)=  0  
  POL(filtersub)=  0  
  POL(neq)=  1  
  POL(true)=  0  
  POL(nil)=  0  
  POL(s)=  0  
  POL(app(x1, x2))=  1 + x2  
  POL(APP(x1, x2))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




The following dependency pairs can be strictly oriented:

APP(app(app(filtersub, false), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)
APP(app(app(filtersub, true), f), app(app(cons, y), ys)) -> APP(app(filter, f), ys)


Additionally, the following usable rules for innermost can be oriented:

app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(filter)=  0  
  POL(0)=  0  
  POL(false)=  0  
  POL(cons)=  0  
  POL(filtersub)=  0  
  POL(neq)=  0  
  POL(true)=  0  
  POL(nil)=  0  
  POL(s)=  0  
  POL(app(x1, x2))=  1 + x2  
  POL(APP(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pairs:

APP(app(filter, app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(filter, f''), ys'')), app(app(filtersub, false), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(app(filtersub, app(f'', y'')), f''), app(app(cons, y''), ys''))), app(filter, f'')), app(app(cons, app(app(cons, y''), ys'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(filter, app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys)) -> APP(app(app(filtersub, app(app(cons, y''), app(app(filter, f''), ys''))), app(app(filtersub, true), f'')), app(app(cons, app(app(cons, y''), ys'')), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




Using the Dependency Graph the DP problem was split into 3 DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 10
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 8
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, app(app(neq, x'), y'')), app(neq, app(s, x'))), app(app(cons, app(s, y'')), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, app(neq, 0)), app(app(cons, app(s, y'')), ys)) -> APP(app(app(filtersub, true), app(neq, 0)), app(app(cons, app(s, y'')), ys))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 10
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 9
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, app(neq, 0)), app(app(cons, 0), ys)) -> APP(app(app(filtersub, false), app(neq, 0)), app(app(cons, 0), ys))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))


Rules:


app(app(neq, 0), 0) -> false
app(app(neq, 0), app(s, y)) -> true
app(app(neq, app(s, x)), 0) -> true
app(app(neq, app(s, x)), app(s, y)) -> app(app(neq, x), y)
app(app(filter, f), nil) -> nil
app(app(filter, f), app(app(cons, y), ys)) -> app(app(app(filtersub, app(f, y)), f), app(app(cons, y), ys))
app(app(app(filtersub, true), f), app(app(cons, y), ys)) -> app(app(cons, y), app(app(filter, f), ys))
app(app(app(filtersub, false), f), app(app(cons, y), ys)) -> app(app(filter, f), ys)
nonzero -> app(filter, app(neq, 0))


Strategy:

innermost




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

APP(app(filter, app(neq, app(s, x'))), app(app(cons, 0), ys)) -> APP(app(app(filtersub, true), app(neq, app(s, x'))), app(app(cons, 0), ys))
no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.


Innermost Termination of R could not be shown.
Duration:
0:08 minutes