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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) -> PLUS(z, s(0))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)


Rules:


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


Strategy:

innermost




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

PLUS(s(x), y) -> PLUS(x, y)
one new Dependency Pair is created:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')


Rules:


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


Strategy:

innermost




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

PLUS(s(s(x'')), y'') -> PLUS(s(x''), y'')
one new Dependency Pair is created:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(s(s(s(x''''))), y'''') -> PLUS(s(s(x'''')), y'''')


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
two new Dependency Pairs are created:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)


Rules:


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


Strategy:

innermost




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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
three new Dependency Pairs are created:

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
one new Dependency Pair is created:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))


Rules:


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


Strategy:

innermost




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

QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))
one new Dependency Pair is created:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 11
                 ↳Argument Filtering and Ordering


Dependency Pairs:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))
QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(x'''''')), s(0), s(0)) -> QUOT(s(x''''''), 0, s(0))


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 14
                 ↳Dependency Graph


Dependency Pair:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 8
                 ↳Narrowing Transformation


Dependency Pairs:

QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
two new Dependency Pairs are created:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 12
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))


Rules:


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


Strategy:

innermost




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

QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
two new Dependency Pairs are created:

QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 13
                 ↳Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(s(x''')))) -> QUOT(x, s(s(plus(x''', s(0)))), s(s(s(x'''))))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(x, 0, s(s(s(s(x''))))) -> QUOT(x, s(s(s(plus(x'', s(0))))), s(s(s(s(x'')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 15
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(x''))))) -> QUOT(x, s(s(s(plus(x'', s(0))))), s(s(s(s(x'')))))
QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(0))) -> QUOT(x, s(s(0)), s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 16
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
QUOT(x, 0, s(s(s(s(x''))))) -> QUOT(x, s(s(s(plus(x'', s(0))))), s(s(s(s(x'')))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(x''''))), s(s(s(y''''))), z'''') -> QUOT(s(s(x'''')), s(s(y'''')), z'''')
two new Dependency Pairs are created:

QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 17
                 ↳Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
QUOT(x, 0, s(s(s(s(x''))))) -> QUOT(x, s(s(s(plus(x'', s(0))))), s(s(s(s(x'')))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(s(s(x''))))) -> QUOT(x, s(s(s(plus(x'', s(0))))), s(s(s(s(x'')))))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 18
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))
QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x'''')), s(s(0)), s(s(x''''''))) -> QUOT(s(x''''), s(0), s(s(x'''''')))
two new Dependency Pairs are created:

QUOT(s(s(x''''0)), s(s(0)), s(s(0))) -> QUOT(s(x''''0), s(0), s(s(0)))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 19
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
QUOT(s(s(x''''0)), s(s(0)), s(s(0))) -> QUOT(s(x''''0), s(0), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(x'''), s(0), s(s(0))) -> QUOT(x''', 0, s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(x''''''''))), s(0), s(s(0))) -> QUOT(s(s(x'''''''')), 0, s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 21
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(x''''''''))), s(0), s(s(0))) -> QUOT(s(s(x'''''''')), 0, s(s(0)))
QUOT(s(s(x''''0)), s(s(0)), s(s(0))) -> QUOT(s(x''''0), s(0), s(s(0)))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x''''0)), s(s(0)), s(s(0))) -> QUOT(s(x''''0), s(0), s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(x'''''''''')))), s(s(0)), s(s(0))) -> QUOT(s(s(s(x''''''''''))), s(0), s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 23
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(x'''''''''')))), s(s(0)), s(s(0))) -> QUOT(s(s(s(x''''''''''))), s(0), s(s(0)))
QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
QUOT(s(s(s(x''''''''))), s(0), s(s(0))) -> QUOT(s(s(x'''''''')), 0, s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x'''''')), 0, s(s(0))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(0))) -> QUOT(s(s(s(s(x'''''''''''')))), s(s(0)), s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 25
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(0))) -> QUOT(s(s(s(s(x'''''''''''')))), s(s(0)), s(s(0)))
QUOT(s(s(s(x''''''''))), s(0), s(s(0))) -> QUOT(s(s(x'''''''')), 0, s(s(0)))
QUOT(s(s(s(s(x'''''''''')))), s(s(0)), s(s(0))) -> QUOT(s(s(s(x''''''''''))), s(0), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(x''''''''))), s(0), s(s(0))) -> QUOT(s(s(x'''''''')), 0, s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(0))) -> QUOT(s(s(s(s(x'''''''''''''')))), 0, s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 27
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(0))) -> QUOT(s(s(s(s(x'''''''''''''')))), 0, s(s(0)))
QUOT(s(s(s(s(x'''''''''')))), s(s(0)), s(s(0))) -> QUOT(s(s(s(x''''''''''))), s(0), s(s(0)))
QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(0))) -> QUOT(s(s(s(s(x'''''''''''')))), s(s(0)), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(x'''''''''')))), s(s(0)), s(s(0))) -> QUOT(s(s(s(x''''''''''))), s(0), s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(s(x'''''''''''''''')))))), s(s(0)), s(s(0))) -> QUOT(s(s(s(s(s(x''''''''''''''''))))), s(0), s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 29
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(s(x'''''''''''''''')))))), s(s(0)), s(s(0))) -> QUOT(s(s(s(s(s(x''''''''''''''''))))), s(0), s(s(0)))
QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(0))) -> QUOT(s(s(s(s(x'''''''''''')))), s(s(0)), s(s(0)))
QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(0))) -> QUOT(s(s(s(s(x'''''''''''''')))), 0, s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(0))) -> QUOT(s(s(s(s(x'''''''''''')))), s(s(0)), s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), 0, s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), s(s(0)), s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 31
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), 0, s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), s(s(0)), s(s(0)))
QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(0))) -> QUOT(s(s(s(s(x'''''''''''''')))), 0, s(s(0)))
QUOT(s(s(s(s(s(s(x'''''''''''''''')))))), s(s(0)), s(s(0))) -> QUOT(s(s(s(s(s(x''''''''''''''''))))), s(0), s(s(0)))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(0))) -> QUOT(s(s(s(s(x'''''''''''''')))), 0, s(s(0)))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(s(s(x''''''''''''''''''''))))))), s(0), s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''''')))))), 0, s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 33
                 ↳Argument Filtering and Ordering


Dependency Pairs:

QUOT(s(s(s(s(s(s(s(x''''''''''''''''''''))))))), s(0), s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''''')))))), 0, s(s(0)))
QUOT(s(s(s(s(s(s(x'''''''''''''''')))))), s(s(0)), s(s(0))) -> QUOT(s(s(s(s(s(x''''''''''''''''))))), s(0), s(s(0)))
QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), 0, s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), s(s(0)), s(s(0)))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

QUOT(s(s(s(s(s(s(s(x''''''''''''''''''''))))))), s(0), s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''''')))))), 0, s(s(0)))
QUOT(s(s(s(s(s(s(x'''''''''''''''')))))), s(s(0)), s(s(0))) -> QUOT(s(s(s(s(s(x''''''''''''''''))))), s(0), s(s(0)))


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 45
                 ↳Dependency Graph


Dependency Pair:

QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), 0, s(s(0))) -> QUOT(s(s(s(s(s(s(x'''''''''''''''''')))))), s(s(0)), s(s(0)))


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 20
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(x'''), s(0), s(s(s(x'''''')))) -> QUOT(x''', 0, s(s(s(x''''''))))
three new Dependency Pairs are created:

QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 22
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(s(0)))) -> QUOT(x, s(s(s(0))), s(s(s(0))))
one new Dependency Pair is created:

QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 24
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(x'''''')))), s(s(s(s(y'''''')))), z'''''') -> QUOT(s(s(s(x''''''))), s(s(s(y''''''))), z'''''')
two new Dependency Pairs are created:

QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 26
                 ↳Narrowing Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(s(s(s(x''')))))) -> QUOT(x, s(s(s(s(plus(x''', s(0)))))), s(s(s(s(s(x'''))))))
two new Dependency Pairs are created:

QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 28
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(x''''''))), s(s(s(0))), s(s(x''''''''))) -> QUOT(s(s(x'''''')), s(s(0)), s(s(x'''''''')))
one new Dependency Pair is created:

QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 30
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))


Rules:


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


Strategy:

innermost




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

QUOT(x, 0, s(s(s(s(0))))) -> QUOT(x, s(s(s(s(0)))), s(s(s(s(0)))))
one new Dependency Pair is created:

QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 32
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x''''0)), s(s(0)), s(s(s(x'''''''')))) -> QUOT(s(x''''0), s(0), s(s(s(x''''''''))))
three new Dependency Pairs are created:

QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 34
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(x''''), s(0), s(s(s(0)))) -> QUOT(x'''', 0, s(s(s(0))))
one new Dependency Pair is created:

QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 35
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(x''''), s(0), s(s(s(s(0))))) -> QUOT(x'''', 0, s(s(s(s(0)))))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 36
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(x'''0), s(0), s(s(s(s(s(x''''')))))) -> QUOT(x'''0, 0, s(s(s(s(s(x'''''))))))
two new Dependency Pairs are created:

QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 37
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(s(x''''''''))))), s(s(s(s(s(y''''''''))))), z'''''''') -> QUOT(s(s(s(s(x'''''''')))), s(s(s(s(y'''''''')))), z'''''''')
two new Dependency Pairs are created:

QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 38
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))
QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(s(x'''''''')))), s(s(s(s(0)))), s(s(x''''''''''))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(x'''''''''')))
one new Dependency Pair is created:

QUOT(s(s(s(s(x''''''''0)))), s(s(s(s(0)))), s(s(s(x'''''''''''')))) -> QUOT(s(s(s(x''''''''0))), s(s(s(0))), s(s(s(x''''''''''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 39
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
QUOT(s(s(s(s(x''''''''0)))), s(s(s(s(0)))), s(s(s(x'''''''''''')))) -> QUOT(s(s(s(x''''''''0))), s(s(s(0))), s(s(s(x''''''''''''))))
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))
QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(s(x'''''''))), s(s(s(0))), s(s(s(x'''''''''')))) -> QUOT(s(s(x''''''')), s(s(0)), s(s(s(x''''''''''))))
three new Dependency Pairs are created:

QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0)))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(0))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(s(0))))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(s(0)))))
QUOT(s(s(s(x'''''''0))), s(s(s(0))), s(s(s(s(s(x''''''''')))))) -> QUOT(s(s(x'''''''0)), s(s(0)), s(s(s(s(s(x'''''''''))))))

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 40
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))
QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0)))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(0))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x''''0')), s(s(0)), s(s(s(0)))) -> QUOT(s(x''''0'), s(0), s(s(s(0))))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(x''''''''''''))))), s(s(0)), s(s(s(0)))) -> QUOT(s(s(s(s(x'''''''''''')))), s(0), s(s(s(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 42
                 ↳Argument Filtering and Ordering


Dependency Pairs:

QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(s(0)), s(s(s(0)))) -> QUOT(s(s(s(s(x'''''''''''')))), s(0), s(s(s(0))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0)))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(0))))
QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))


Rules:


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


Strategy:

innermost




The following dependency pairs can be strictly oriented:

QUOT(s(s(s(s(x'''''''''')))), s(0), s(s(s(0)))) -> QUOT(s(s(s(x''''''''''))), 0, s(s(s(0))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(s(0)), s(s(s(0)))) -> QUOT(s(s(s(s(x'''''''''''')))), s(0), s(s(s(0))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0)))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(0))))


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 46
                 ↳Dependency Graph


Dependency Pair:

QUOT(s(s(s(x''''''''))), 0, s(s(s(0)))) -> QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(0))))


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 41
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(x'''''''0))), s(s(s(0))), s(s(s(s(s(x''''''''')))))) -> QUOT(s(s(x'''''''0)), s(s(0)), s(s(s(s(s(x'''''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(s(0))))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(s(0)))))
QUOT(s(s(s(s(x''''''''0)))), s(s(s(s(0)))), s(s(s(x'''''''''''')))) -> QUOT(s(s(s(x''''''''0))), s(s(s(0))), s(s(s(x''''''''''''))))
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(0))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(0)))))
one new Dependency Pair is created:

QUOT(s(s(s(s(s(s(x'''''''''''''')))))), s(s(0)), s(s(s(s(0))))) -> QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(s(s(0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 43
                 ↳Forward Instantiation Transformation


Dependency Pairs:

QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
QUOT(s(s(s(x'''''''0))), s(s(s(0))), s(s(s(s(s(x''''''''')))))) -> QUOT(s(s(x'''''''0)), s(s(0)), s(s(s(s(s(x'''''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(x'''''''''''''')))))), s(s(0)), s(s(s(s(0))))) -> QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(s(0))))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(s(0)))))
QUOT(s(s(s(s(x''''''''0)))), s(s(s(s(0)))), s(s(s(x'''''''''''')))) -> QUOT(s(s(s(x''''''''0))), s(s(s(0))), s(s(s(x''''''''''''))))
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))
QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')


Rules:


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


Strategy:

innermost




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

QUOT(s(s(x''''0')), s(s(0)), s(s(s(s(s(x''''''')))))) -> QUOT(s(x''''0'), s(0), s(s(s(s(s(x'''''''))))))
two new Dependency Pairs are created:

QUOT(s(s(x''''0'')), s(s(0)), s(s(s(s(s(0)))))) -> QUOT(s(x''''0''), s(0), s(s(s(s(s(0))))))
QUOT(s(s(x''''0'')), s(s(0)), s(s(s(s(s(s(x''''''))))))) -> QUOT(s(x''''0''), s(0), s(s(s(s(s(s(x'''''')))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 44
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

QUOT(s(x'''0'), s(0), s(s(s(s(s(s(x''''))))))) -> QUOT(x'''0', 0, s(s(s(s(s(s(x'''')))))))
QUOT(s(s(x''''0'')), s(s(0)), s(s(s(s(s(s(x''''''))))))) -> QUOT(s(x''''0''), s(0), s(s(s(s(s(s(x'''''')))))))
QUOT(x, 0, s(s(s(s(s(0)))))) -> QUOT(x, s(s(s(s(s(0))))), s(s(s(s(s(0))))))
QUOT(s(x'''0'), s(0), s(s(s(s(s(0)))))) -> QUOT(x'''0', 0, s(s(s(s(s(0))))))
QUOT(s(s(x''''0'')), s(s(0)), s(s(s(s(s(0)))))) -> QUOT(s(x''''0''), s(0), s(s(s(s(s(0))))))
QUOT(s(s(s(x'''''''0))), s(s(s(0))), s(s(s(s(s(x''''''''')))))) -> QUOT(s(s(x'''''''0)), s(s(0)), s(s(s(s(s(x'''''''''))))))
QUOT(s(s(s(s(x'''''''''')))), 0, s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(x''''''''''''))))), s(0), s(s(s(s(0))))) -> QUOT(s(s(s(s(x'''''''''''')))), 0, s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(x'''''''''''''')))))), s(s(0)), s(s(s(s(0))))) -> QUOT(s(s(s(s(s(x''''''''''''''))))), s(0), s(s(s(s(0)))))
QUOT(s(s(s(x''''''''))), s(s(s(0))), s(s(s(s(0))))) -> QUOT(s(s(x'''''''')), s(s(0)), s(s(s(s(0)))))
QUOT(s(s(s(s(x''''''''0)))), s(s(s(s(0)))), s(s(s(x'''''''''''')))) -> QUOT(s(s(s(x''''''''0))), s(s(s(0))), s(s(s(x''''''''''''))))
QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(0))))), s(s(x''''''''''''))) -> QUOT(s(s(s(s(x'''''''''')))), s(s(s(s(0)))), s(s(x'''''''''''')))
QUOT(s(s(s(s(s(s(x'''''''''')))))), s(s(s(s(s(s(y'''''''''')))))), z'''''''''') -> QUOT(s(s(s(s(s(x''''''''''))))), s(s(s(s(s(y''''''''''))))), z'''''''''')
QUOT(x, 0, s(s(s(s(s(s(x''))))))) -> QUOT(x, s(s(s(s(s(plus(x'', s(0))))))), s(s(s(s(s(s(x'')))))))


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:03 minutes