Term Rewriting System R:
[f, x, y, z]
app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
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(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
APP(app(times, app(s, x)), y) -> APP(plus, app(app(times, x), y))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(times, app(s, x)), y) -> APP(times, x)
SUM -> APP(app(fold, add), 0)
SUM -> APP(fold, add)
PROD -> APP(app(fold, mul), app(s, 0))
PROD -> APP(fold, mul)
PROD -> APP(s, 0)

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, x)), y) -> APP(app(plus, app(app(times, x), y)), y)
two new Dependency Pairs are created:

APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')
APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, 0)), y'') -> APP(app(plus, 0), y'')
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 3
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, x''))), y'') -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), y'')), y'')
two new Dependency Pairs are created:

APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, app(app(plus, 0), y''')), y''')
APP(app(times, app(s, app(s, app(s, x')))), y''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''')), y''')), y''')), y''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 4
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, app(s, x')))), y''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''')), y''')), y''')), y''')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, app(app(plus, 0), y''')), y''')
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, app(app(plus, 0), y''')), y''')
one new Dependency Pair is created:

APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 5
                 ↳Forward Instantiation Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, app(s, x')))), y''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''')), y''')), y''')), y''')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(f, y)
five new Dependency Pairs are created:

APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 6
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, app(s, x')))), y''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''')), y''')), y''')), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, x')))), y''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''')), y''')), y''')), y''')
two new Dependency Pairs are created:

APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, 0), y'''')), y'''')), y'''')
APP(app(times, app(s, app(s, app(s, app(s, x''))))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''')), y'''')), y'''')), y'''')), y'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 7
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, app(s, app(s, x''))))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''')), y'''')), y'''')), y'''')), y'''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, 0), y'''')), y'''')), y'''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, 0), y'''')), y'''')), y'''')
one new Dependency Pair is created:

APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, y''''), y'''')), y'''')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, y''''), y'''')), y'''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, app(s, app(s, x''))))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''')), y'''')), y'''')), y'''')), y'''')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, app(s, x''))))), y'''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''')), y'''')), y'''')), y'''')), y'''')
two new Dependency Pairs are created:

APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y''''')), y''''')), y''''')), y''''')
APP(app(times, app(s, app(s, app(s, app(s, app(s, x')))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''''')), y''''')), y''''')), y''''')), y''''')), y''''')

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

APP(app(times, app(s, app(s, app(s, app(s, app(s, x')))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''''')), y''''')), y''''')), y''''')), y''''')), y''''')
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y''''')), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, y''''), y'''')), y'''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y''''')), y''''')), y''''')), y''''')
one new Dependency Pair is created:

APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 10
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, y''''), y'''')), y'''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, app(s, app(s, app(s, x')))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''''')), y''''')), y''''')), y''''')), y''''')), y''''')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, 0)))), y'''') -> APP(app(plus, app(app(plus, y''''), y'''')), y'''')
two new Dependency Pairs are created:

APP(app(times, app(s, app(s, app(s, 0)))), 0) -> APP(app(plus, 0), 0)
APP(app(times, app(s, app(s, app(s, 0)))), app(s, x')) -> APP(app(plus, app(s, app(app(plus, x'), app(s, x')))), app(s, x'))

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 Pairs:

APP(app(times, app(s, app(s, app(s, 0)))), app(s, x')) -> APP(app(plus, app(s, app(app(plus, x'), app(s, x')))), app(s, x'))
APP(app(times, app(s, app(s, app(s, 0)))), 0) -> APP(app(plus, 0), 0)
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, app(s, app(s, x')))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''''')), y''''')), y''''')), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, app(s, app(s, x')))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x'), y''''')), y''''')), y''''')), y''''')), y''''')), y''''')
two new Dependency Pairs are created:

APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(times, app(s, app(s, app(s, app(s, app(s, app(s, x''))))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 12
                 ↳Rewriting Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, app(s, app(s, app(s, app(s, x''))))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(times, app(s, app(s, app(s, 0)))), 0) -> APP(app(plus, 0), 0)
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(times, app(s, app(s, app(s, 0)))), app(s, x')) -> APP(app(plus, app(s, app(app(plus, x'), app(s, x')))), app(s, x'))


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, 0), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')
one new Dependency Pair is created:

APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, y''''''), y'''''')), y'''''')), y'''''')), y'''''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 13
                 ↳Narrowing Transformation


Dependency Pairs:

APP(app(times, app(s, app(s, app(s, 0)))), app(s, x')) -> APP(app(plus, app(s, app(app(plus, x'), app(s, x')))), app(s, x'))
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, y''''''), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(times, app(s, app(s, app(s, 0)))), 0) -> APP(app(plus, 0), 0)
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(times, app(s, app(s, app(s, app(s, app(s, app(s, x''))))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost




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

APP(app(times, app(s, app(s, app(s, 0)))), 0) -> APP(app(plus, 0), 0)
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 14
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

APP(app(times, app(s, app(s, app(s, app(s, app(s, 0)))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, y''''''), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(times, app(s, app(s, app(s, app(s, app(s, app(s, x''))))))), y'''''') -> APP(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(plus, app(app(times, x''), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')), y'''''')
APP(app(app(fold, app(times, app(s, app(s, 0)))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, 0))), y')
APP(app(times, app(s, app(s, app(s, app(s, 0))))), y''''') -> APP(app(plus, app(app(plus, app(app(plus, y'''''), y''''')), y''''')), y''''')
APP(app(app(fold, app(times, app(s, app(s, app(s, x'''))))), x), app(app(cons, y'), z)) -> APP(app(times, app(s, app(s, app(s, x''')))), y')
APP(app(times, app(s, app(s, 0))), y''') -> APP(app(plus, y'''), y''')
APP(app(times, app(s, x)), y) -> APP(app(times, x), y)
APP(app(app(fold, app(times, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(times, app(s, x'')), y'')
APP(app(app(fold, app(plus, app(s, x''))), x), app(app(cons, y''), z)) -> APP(app(plus, app(s, x'')), y'')
APP(app(app(fold, app(app(fold, f''), x'')), x), app(app(cons, app(app(cons, y''), z'')), z)) -> APP(app(app(fold, f''), x''), app(app(cons, y''), z''))
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(app(fold, f), x), z)
APP(app(app(fold, f), x), app(app(cons, y), z)) -> APP(app(f, y), app(app(app(fold, f), x), z))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(times, app(s, app(s, app(s, 0)))), app(s, x')) -> APP(app(plus, app(s, app(app(plus, x'), app(s, x')))), app(s, x'))


Rules:


app(app(app(fold, f), x), nil) -> x
app(app(app(fold, f), x), app(app(cons, y), z)) -> app(app(f, y), app(app(app(fold, f), x), z))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(times, 0), y) -> 0
app(app(times, app(s, x)), y) -> app(app(plus, app(app(times, x), y)), y)
sum -> app(app(fold, add), 0)
prod -> app(app(fold, mul), app(s, 0))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:04 minutes