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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

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
                 ↳Argument Filtering and Ordering


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




The following dependency pairs can be strictly oriented:

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


The following usable rules for innermost can be oriented:

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


Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
plus > s

resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2, x3) -> x1
s(x1) -> s(x1)
plus(x1, x2) -> plus(x1, x2)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳FwdInst
             ...
               →DP Problem 15
                 ↳Dependency Graph


Dependency Pairs:

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




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes