Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

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)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
IFQUOT(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →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)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


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 4
Forward Instantiation Transformation
       →DP Problem 2
FwdInst
       →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)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


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 4
FwdInst
             ...
               →DP Problem 5
Polynomial Ordering
       →DP Problem 2
FwdInst
       →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)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


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 w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(MINUS(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
           →DP Problem 4
FwdInst
             ...
               →DP Problem 6
Dependency Graph
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



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


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 3
Nar


Dependency Pair:

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


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LE(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 9
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

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

QUOT(0, s(y')) -> IFQUOT(false, 0, s(y'))
QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Narrowing Transformation


Dependency Pairs:

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 11
Narrowing Transformation


Dependency Pairs:

IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))
three new Dependency Pairs are created:

QUOT(s(y'''), s(0)) -> IFQUOT(true, s(y'''), s(0))
QUOT(s(0), s(s(x'))) -> IFQUOT(false, s(0), s(s(x')))
QUOT(s(s(y')), s(s(x'))) -> IFQUOT(le(x', y'), s(s(y')), s(s(x')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 12
Narrowing Transformation


Dependency Pairs:

QUOT(s(s(y')), s(s(x'))) -> IFQUOT(le(x', y'), s(s(y')), s(s(x')))
QUOT(s(y'''), s(0)) -> IFQUOT(true, s(y'''), s(0))
IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

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

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

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 13
Polynomial Ordering


Dependency Pairs:

IFQUOT(true, s(s(x')), s(s(y'))) -> QUOT(minus(x', y'), s(s(y')))
QUOT(s(s(y')), s(s(x'))) -> IFQUOT(le(x', y'), s(s(y')), s(s(x')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFQUOT(true, s(s(x')), s(s(y'))) -> QUOT(minus(x', y'), s(s(y')))


Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  x1  
  POL(0)=  0  
  POL(false)=  0  
  POL(minus(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(IF_QUOT(x1, x2, x3))=  x2  
  POL(le(x1, x2))=  0  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 15
Dependency Graph


Dependency Pair:

QUOT(s(s(y')), s(s(x'))) -> IFQUOT(le(x', y'), s(s(y')), s(s(x')))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 14
Polynomial Ordering


Dependency Pairs:

IFQUOT(true, s(x'''), s(0)) -> QUOT(x''', s(0))
QUOT(s(y'''), s(0)) -> IFQUOT(true, s(y'''), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  x1  
  POL(0)=  0  
  POL(true)=  0  
  POL(IF_QUOT(x1, x2, x3))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Nar
           →DP Problem 10
Nar
             ...
               →DP Problem 16
Dependency Graph


Dependency Pair:

QUOT(s(y'''), s(0)) -> IFQUOT(true, s(y'''), s(0))


Rules:


minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes