Term Rewriting System R:
[x, y]
app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(minus, x), app(s, y)) -> APP(pred, app(app(minus, x), y))
APP(app(minus, x), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(minus, x)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(pred)=  0  
  POL(minus)=  0  
  POL(quot)=  0  
  POL(s)=  1  
  POL(app(x1, x2))=  x1 + x2  
  POL(APP(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

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

APP(app(quot, app(s, x'')), app(s, 0)) -> APP(app(quot, x''), app(s, 0))
APP(app(quot, app(s, x'')), app(s, app(s, y''))) -> APP(app(quot, app(pred, app(app(minus, x''), y''))), app(s, app(s, y'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Narrowing Transformation


Dependency Pairs:

APP(app(quot, app(s, x'')), app(s, app(s, y''))) -> APP(app(quot, app(pred, app(app(minus, x''), y''))), app(s, app(s, y'')))
APP(app(quot, app(s, x'')), app(s, 0)) -> APP(app(quot, x''), app(s, 0))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
no new Dependency Pairs are created.
The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 5
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x'')), app(s, app(s, y''))) -> APP(app(quot, app(pred, app(app(minus, x''), y''))), app(s, app(s, y'')))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x'')), app(s, app(s, y''))) -> APP(app(quot, app(pred, app(app(minus, x''), y''))), app(s, app(s, y'')))
two new Dependency Pairs are created:

APP(app(quot, app(s, x''')), app(s, app(s, 0))) -> APP(app(quot, app(pred, x''')), app(s, app(s, 0)))
APP(app(quot, app(s, x''')), app(s, app(s, app(s, y')))) -> APP(app(quot, app(pred, app(pred, app(app(minus, x'''), y')))), app(s, app(s, app(s, y'))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x''')), app(s, app(s, 0))) -> APP(app(quot, app(pred, x''')), app(s, app(s, 0)))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x''')), app(s, app(s, 0))) -> APP(app(quot, app(pred, x''')), app(s, app(s, 0)))
one new Dependency Pair is created:

APP(app(quot, app(s, app(s, x'))), app(s, app(s, 0))) -> APP(app(quot, x'), app(s, app(s, 0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 8
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x''')), app(s, app(s, app(s, y')))) -> APP(app(quot, app(pred, app(pred, app(app(minus, x'''), y')))), app(s, app(s, app(s, y'))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x''')), app(s, app(s, app(s, y')))) -> APP(app(quot, app(pred, app(pred, app(app(minus, x'''), y')))), app(s, app(s, app(s, y'))))
two new Dependency Pairs are created:

APP(app(quot, app(s, x'''')), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, app(pred, x''''))), app(s, app(s, app(s, 0))))
APP(app(quot, app(s, x'''')), app(s, app(s, app(s, app(s, y''))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(app(minus, x''''), y''))))), app(s, app(s, app(s, app(s, y'')))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 10
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x'''')), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, app(pred, x''''))), app(s, app(s, app(s, 0))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x'''')), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, app(pred, x''''))), app(s, app(s, app(s, 0))))
one new Dependency Pair is created:

APP(app(quot, app(s, app(s, x'))), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, x')), app(s, app(s, app(s, 0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 12
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, app(s, x'))), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, x')), app(s, app(s, app(s, 0))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, app(s, x'))), app(s, app(s, app(s, 0)))) -> APP(app(quot, app(pred, x')), app(s, app(s, app(s, 0))))
one new Dependency Pair is created:

APP(app(quot, app(s, app(s, app(s, x'')))), app(s, app(s, app(s, 0)))) -> APP(app(quot, x''), app(s, app(s, app(s, 0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 11
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x'''')), app(s, app(s, app(s, app(s, y''))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(app(minus, x''''), y''))))), app(s, app(s, app(s, app(s, y'')))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x'''')), app(s, app(s, app(s, app(s, y''))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(app(minus, x''''), y''))))), app(s, app(s, app(s, app(s, y'')))))
two new Dependency Pairs are created:

APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, 0))))) -> APP(app(quot, app(pred, app(pred, app(pred, x''''')))), app(s, app(s, app(s, app(s, 0)))))
APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, app(s, y')))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(pred, app(app(minus, x'''''), y')))))), app(s, app(s, app(s, app(s, app(s, y'))))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 13
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, 0))))) -> APP(app(quot, app(pred, app(pred, app(pred, x''''')))), app(s, app(s, app(s, app(s, 0)))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, 0))))) -> APP(app(quot, app(pred, app(pred, app(pred, x''''')))), app(s, app(s, app(s, app(s, 0)))))
one new Dependency Pair is created:

APP(app(quot, app(s, app(s, x'))), app(s, app(s, app(s, app(s, 0))))) -> APP(app(quot, app(pred, app(pred, x'))), app(s, app(s, app(s, app(s, 0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 14
                 ↳Narrowing Transformation


Dependency Pair:

APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, app(s, y')))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(pred, app(app(minus, x'''''), y')))))), app(s, app(s, app(s, app(s, app(s, y'))))))


Rules:


app(pred, app(s, x)) -> x
app(app(minus, x), 0) -> x
app(app(minus, x), app(s, y)) -> app(pred, app(app(minus, x), y))
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))


Strategy:

innermost




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

APP(app(quot, app(s, x''''')), app(s, app(s, app(s, app(s, app(s, y')))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(pred, app(app(minus, x'''''), y')))))), app(s, app(s, app(s, app(s, app(s, y'))))))
two new Dependency Pairs are created:

APP(app(quot, app(s, x'''''')), app(s, app(s, app(s, app(s, app(s, 0)))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(pred, x''''''))))), app(s, app(s, app(s, app(s, app(s, 0))))))
APP(app(quot, app(s, x'''''')), app(s, app(s, app(s, app(s, app(s, app(s, y''))))))) -> APP(app(quot, app(pred, app(pred, app(pred, app(pred, app(pred, app(app(minus, x''''''), y''))))))), app(s, app(s, app(s, app(s, app(s, app(s, y'')))))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:01 minutes