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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x)
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)
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(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(plus, app(app(minus, y), app(s, app(s, z))))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(plus, app(app(plus, y), app(s, app(s, z))))

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

APP(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))
APP(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
APP(app(plus, app(s, x)), y) -> APP(app(plus, 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(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)


Rules:


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


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> 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)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), 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(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 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 3
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> 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)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), 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(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> APP(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 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 4
                 ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> 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)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), 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(quot, app(s, app(s, x''))), app(s, app(s, y''))) -> APP(app(quot, app(app(minus, x''), y'')), app(s, app(s, y'')))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> 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)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), 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(quot, app(s, app(s, app(s, x')))), app(s, app(s, app(s, y')))) -> APP(app(quot, app(app(minus, x'), y')), app(s, app(s, app(s, y'))))
two new Dependency Pairs are 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))))
APP(app(quot, app(s, app(s, app(s, app(s, x''))))), app(s, app(s, app(s, app(s, y''))))) -> APP(app(quot, app(app(minus, x''), y'')), app(s, app(s, app(s, app(s, 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(quot, app(s, app(s, app(s, app(s, x''))))), app(s, app(s, app(s, app(s, y''))))) -> APP(app(quot, app(app(minus, x''), y'')), app(s, app(s, app(s, app(s, y'')))))
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))))
APP(app(quot, app(s, x'')), app(s, 0)) -> APP(app(quot, x''), app(s, 0))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, app(s, x'''))), app(s, app(s, 0))) -> APP(app(quot, x'''), app(s, app(s, 0)))


Rules:


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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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


Rules:


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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Nar
           →DP Problem 2
             ↳Nar
             ...
               →DP Problem 8
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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


Rules:


app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> 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)))
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(app(plus, app(app(minus, x), app(s, 0))), app(app(minus, y), app(s, app(s, z)))) -> app(app(plus, app(app(minus, y), app(s, app(s, z)))), app(app(minus, x), app(s, 0)))
app(app(plus, app(app(plus, x), app(s, 0))), app(app(plus, y), app(s, app(s, z)))) -> app(app(plus, app(app(plus, y), app(s, app(s, z)))), app(app(plus, x), app(s, 0)))


Strategy:

innermost



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