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

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

Furthermore, R contains three SCCs.


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


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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 4
Forward Instantiation Transformation
       →DP Problem 2
Nar
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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 4
FwdInst
             ...
               →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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 4
FwdInst
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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
       →DP Problem 3
Nar


Dependency Pairs:

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
MINUS(s(x), s(y)) -> MINUS(x, y)


Rules:


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

MINUS(minus(x, y), z) -> MINUS(x, plus(y, z))
one new Dependency Pair is created:

MINUS(minus(x, s(x'')), z') -> MINUS(x, s(plus(x'', z')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(s(s(x''''))), s(s(s(y'''')))) -> MINUS(s(s(x'''')), s(s(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:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
FwdInst
             ...
               →DP Problem 10
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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 3
Narrowing Transformation


Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(x), s(y)) -> QUOT(minus(x, y), s(y))
three new Dependency Pairs are created:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(s(x'')), s(s(y''))) -> QUOT(minus(x'', y''), s(s(y'')))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(x'')), s(s(y''))) -> QUOT(minus(x'', y''), s(s(y'')))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))
QUOT(s(x''), s(0)) -> QUOT(x'', s(0))


Rules:


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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 12
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', y'')), s(y0)) -> QUOT(minus(x'', plus(y'', y0)), s(y0))
one new Dependency Pair is created:

QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 13
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))


Rules:


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

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', 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 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 14
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(s(x'))), s(s(s(y')))) -> QUOT(minus(x', y'), s(s(s(y'))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))


Rules:


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

QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 15
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(minus(x', y'))), s(s(y'''))) -> QUOT(minus(x', plus(y', y''')), s(s(y''')))
one new Dependency Pair is created:

QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 16
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(x'))), s(y0')) -> QUOT(minus(x'', s(plus(x', y0'))), s(y0'))
two new Dependency Pairs are created:

QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 17
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(s(s(x'')))), s(s(s(s(y''))))) -> QUOT(minus(x'', y''), s(s(s(s(y'')))))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))


Rules:


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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 18
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(s(minus(x'', y0)))), s(s(s(y'')))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(y''))))
one new Dependency Pair is created:

QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 19
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(x'')))), s(s(y''''))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(y'''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))


Rules:


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

QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 20
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(0))), s(y0'')) -> QUOT(minus(x'', s(y0'')), s(y0''))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 21
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(s(x0)))), s(y0'')) -> QUOT(minus(x'', s(s(plus(x0, y0'')))), s(y0''))
two new Dependency Pairs are created:

QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 22
Narrowing Transformation


Dependency Pairs:

QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(s(s(x'))))), s(s(s(s(s(y')))))) -> QUOT(minus(x', y'), s(s(s(s(s(y'))))))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))


Rules:


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

QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 23
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(s(s(minus(x', y'))))), s(s(s(s(y'''))))) -> QUOT(minus(x', plus(y', y''')), s(s(s(s(y''')))))
one new Dependency Pair is created:

QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 24
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(x'))))), s(s(s(y''')))) -> QUOT(minus(x'', s(plus(x', y'''))), s(s(s(y'''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))


Rules:


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

QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 25
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(minus(x', s(0)))), s(s(y'''''))) -> QUOT(minus(x', s(y''''')), s(s(y''''')))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 26
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(minus(x', s(s(x0))))), s(s(y'''''))) -> QUOT(minus(x', s(s(plus(x0, y''''')))), s(s(y''''')))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 27
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(s(0)))), s(y0''')) -> QUOT(minus(x'', s(s(y0'''))), s(y0'''))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 28
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(s(s(x'))))), s(y0''')) -> QUOT(minus(x'', s(s(s(plus(x', y0'''))))), s(y0'''))
two new Dependency Pairs are created:

QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 29
Narrowing Transformation


Dependency Pairs:

QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(s(minus(x'', s(0))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(y'''')), s(s(s(y''''))))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 30
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(s(minus(x', s(s(0))))), s(s(y''''''))) -> QUOT(minus(x', s(s(y''''''))), s(s(y'''''')))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 31
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
minus(minus(x, y), z) -> minus(x, plus(y, z))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
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(s(minus(x'', s(s(s(0))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(y0'''')))), s(y0''''))
no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 32
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(x''')), s(s(0))) -> QUOT(x''', s(s(0)))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))


Rules:


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

QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 33
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))


Rules:


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

QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 34
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(x''))), s(s(s(0)))) -> QUOT(x'', s(s(s(0))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))


Rules:


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

QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), 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 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 35
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(x''')))), s(s(s(s(0))))) -> QUOT(x''', s(s(s(s(0)))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))


Rules:


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

QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), 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 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 36
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(s(s(x''''')))), s(s(0))) -> QUOT(s(s(x''''')), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))


Rules:


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

QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 37
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(x''''''))), s(0)) -> QUOT(s(s(x'''''')), s(0))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))


Rules:


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

QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 38
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(x'''')))))), s(s(s(0)))) -> QUOT(s(s(s(x''''))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))


Rules:


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

QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), 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 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 39
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(x''''''')))))), s(s(0))) -> QUOT(s(s(s(s(x''''''')))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))


Rules:


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

QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 40
Forward Instantiation Transformation


Dependency Pairs:

QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))
QUOT(s(s(s(s(x'''''''')))), s(0)) -> QUOT(s(s(s(x''''''''))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))


Rules:


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

QUOT(s(s(s(s(minus(x''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(s(x''''''''''))))), s(0)) -> QUOT(s(s(s(s(x'''''''''')))), s(0))
QUOT(s(s(s(s(s(minus(x'''''''''', s(x'0'''''''))))))), s(0)) -> QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''''')))))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(s(s(s(x0''''''))))))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(s(s(s(x0'''''')))))))), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 11
Nar
             ...
               →DP Problem 41
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

QUOT(s(s(s(s(minus(x'''''''''', s(s(s(s(x0''''''))))))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(s(s(s(x0'''''')))))))), s(0))
QUOT(s(s(s(s(s(minus(x'''''''''', s(x'0'''''''))))))), s(0)) -> QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''''')))))), s(0))
QUOT(s(s(s(s(s(x''''''''''))))), s(0)) -> QUOT(s(s(s(s(x'''''''''')))), s(0))
QUOT(s(s(s(s(minus(x'''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x''''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x''''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(s(minus(x'''''''', s(x'0''''')))))), s(0)) -> QUOT(s(s(s(minus(x'''''''', s(x'0'''''))))), s(0))
QUOT(s(s(s(minus(x''''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x''''''', s(x'0''')))), s(0))
QUOT(s(s(minus(x''''', s(s(s(s(x0''))))))), s(0)) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(0))
QUOT(s(s(s(minus(x'''''', s(x'0'''))))), s(0)) -> QUOT(s(s(minus(x'''''', s(x'0''')))), s(0))
QUOT(s(s(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(minus(x''''''''', s(s(s(s(x0'''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(minus(x'0''''', s(s(s(x''''''''')))))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(s(x''''''''')))))))), s(s(0))) -> QUOT(s(s(s(s(s(s(x''''''''')))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'''0'', s(s(s(x''''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x'''''''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(s(minus(x'''''', s(s(s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(minus(x'''''', s(s(s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(minus(x'''0'', s(s(s(x''''''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(minus(x'''0'', s(s(s(x'''''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(minus(x'''''', s(s(x0'''')))))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(minus(x'''''', s(s(x0''''))))))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(s(s(s(x''''''))))))))), s(s(s(0)))) -> QUOT(s(s(s(s(s(s(x'''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x''''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))))), s(s(0))) -> QUOT(s(s(s(minus(x''''''', s(s(s(s(x0'''')))))))), s(s(0)))
QUOT(s(s(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))))), s(s(0))) -> QUOT(s(s(s(s(minus(x'0''', s(s(s(x''''''')))))))), s(s(0)))
QUOT(s(s(s(s(minus(x'''0, s(s(s(x'''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'''0, s(s(s(x'''''')))))), s(s(0)))
QUOT(s(s(s(s(s(minus(x''''', s(s(s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(minus(x'0', s(s(s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(minus(x''''', s(s(x0'')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(minus(x''''', s(s(x0'')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(minus(x'0', s(x''''')))))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(minus(x'0', s(x''''')))))), s(s(s(s(0)))))
QUOT(s(s(s(s(s(s(s(s(x''''')))))))), s(s(s(s(0))))) -> QUOT(s(s(s(s(x''''')))), s(s(s(s(0)))))
QUOT(s(s(s(s(minus(x'''', s(s(s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(minus(x'''', s(s(s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(s(s(minus(x'''0, s(s(s(x'''''))))))))), s(s(s(0)))) -> QUOT(s(s(minus(x'''0, s(s(s(x''''')))))), s(s(s(0))))
QUOT(s(s(s(s(s(s(minus(x'''', s(s(x0''))))))))), s(s(s(0)))) -> QUOT(s(s(s(minus(x'''', s(s(x0'')))))), s(s(s(0))))
QUOT(s(s(s(minus(x''''', s(s(s(s(x0'')))))))), s(s(0))) -> QUOT(s(minus(x''''', s(s(s(s(x0'')))))), s(s(0)))
QUOT(s(s(s(s(minus(x'0', s(s(s(x''''')))))))), s(s(0))) -> QUOT(s(s(minus(x'0', s(s(s(x''''')))))), s(s(0)))
QUOT(s(s(minus(x', s(s(s(x'')))))), s(s(y''''''))) -> QUOT(minus(x', s(s(s(plus(x'', y''''''))))), s(s(y'''''')))
QUOT(s(s(s(minus(x'', s(s(x0)))))), s(s(s(y'''')))) -> QUOT(minus(x'', s(s(plus(x0, y'''')))), s(s(s(y''''))))
QUOT(s(s(s(s(minus(x', s(x'')))))), s(s(s(s(y''''))))) -> QUOT(minus(x', s(plus(x'', y''''))), s(s(s(s(y'''')))))
QUOT(s(s(s(s(s(s(x'')))))), s(s(s(s(s(s(y''))))))) -> QUOT(minus(x'', y''), s(s(s(s(s(s(y'')))))))
QUOT(s(s(s(s(s(minus(x'', y0)))))), s(s(s(s(s(y'')))))) -> QUOT(minus(x'', plus(y0, y'')), s(s(s(s(s(y''))))))
QUOT(s(s(s(s(s(x''))))), s(s(s(s(s(0)))))) -> QUOT(x'', s(s(s(s(s(0))))))
QUOT(s(minus(x'', s(s(s(s(x0)))))), s(y0'''')) -> QUOT(minus(x'', s(s(s(s(plus(x0, y0'''')))))), s(y0''''))
QUOT(s(s(minus(x'''', s(x'0')))), s(0)) -> QUOT(s(minus(x'''', s(x'0'))), s(0))
QUOT(s(s(s(minus(x'''''''', s(s(s(s(x0'''')))))))), s(0)) -> QUOT(s(s(minus(x'''''''', s(s(s(s(x0''''))))))), s(0))


Rules:


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


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:52 minutes