Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(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:

LE(s(X), s(Y)) -> LE(X, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


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 4
Forward Instantiation Transformation
       →DP Problem 2
Nar
       →DP Problem 3
Nar


Dependency Pair:

LE(s(s(X'')), s(s(Y''))) -> LE(s(X''), s(Y''))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


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 4
FwdInst
             ...
               →DP Problem 5
Argument Filtering and Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Nar


Dependency Pair:

LE(s(s(s(X''''))), s(s(s(Y'''')))) -> LE(s(s(X'''')), s(s(Y'''')))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


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 that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LE(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(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:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
two new Dependency Pairs are created:

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), 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
Narrowing Transformation
       →DP Problem 3
Nar


Dependency Pairs:

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
three new Dependency Pairs are created:

MINUS(s(0), s(Y''')) -> IFMINUS(true, s(0), s(Y'''))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), 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
Nar
             ...
               →DP Problem 8
Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pairs:

MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
three new Dependency Pairs are created:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))
IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))

The transformation is resulting in three new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

IFMINUS(false, s(s(X''')), s(s(Y'''))) -> MINUS(s(X'''), s(s(Y''')))
one new Dependency Pair is created:

IFMINUS(false, s(s(s(X''''))), 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
Nar
             ...
               →DP Problem 12
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(s(s(X''''))), s(s(Y''''))) -> MINUS(s(s(X'''')), s(s(Y'''')))
MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(s(X''''))), s(s(Y''''))) -> MINUS(s(s(X'''')), s(s(Y'''')))


The following usable rules for innermost can be oriented:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)


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

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 16
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:

MINUS(s(s(X')), s(s(Y'))) -> IFMINUS(le(X', Y'), s(s(X')), s(s(Y')))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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 7
Nar
             ...
               →DP Problem 10
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

IFMINUS(false, s(s(X''')), s(0)) -> MINUS(s(X'''), s(0))
one new Dependency Pair is created:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 13
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))
MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(s(X')), s(0)) -> IFMINUS(false, s(s(X')), s(0))


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(IFMINUS(x1, x2, x3))=  x1 + x2 + x3  
  POL(false)=  0  
  POL(MINUS(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 17
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:

IFMINUS(false, s(s(s(X''''))), s(0)) -> MINUS(s(s(X'''')), s(0))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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 7
Nar
             ...
               →DP Problem 11
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 14
Forward Instantiation Transformation
       →DP Problem 3
Nar


Dependency Pairs:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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''), 0) -> IFMINUS(false, s(X''), 0)
one new Dependency Pair is created:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 15
Argument Filtering and Ordering
       →DP Problem 3
Nar


Dependency Pairs:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)
IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)


There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(IFMINUS(x1, x2, x3))=  x1 + x2 + x3  
  POL(false)=  0  
  POL(MINUS(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
           →DP Problem 7
Nar
             ...
               →DP Problem 18
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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))
two new Dependency Pairs are created:

QUOT(s(0), s(Y'')) -> QUOT(0, s(Y''))
QUOT(s(s(X'')), s(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), 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 19
Narrowing Transformation


Dependency Pair:

QUOT(s(s(X'')), s(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), s(Y''))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(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(Y'')) -> QUOT(ifMinus(le(s(X''), Y''), s(X''), Y''), s(Y''))
two new Dependency Pairs are created:

QUOT(s(s(X''')), s(0)) -> QUOT(ifMinus(false, s(X'''), 0), s(0))
QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 20
Rewriting Transformation


Dependency Pair:

QUOT(s(s(X''')), s(0)) -> QUOT(ifMinus(false, s(X'''), 0), s(0))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




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

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

QUOT(s(s(X''')), s(0)) -> QUOT(s(minus(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 19
Nar
             ...
               →DP Problem 22
Argument Filtering and Ordering


Dependency Pair:

QUOT(s(s(X''')), s(0)) -> QUOT(s(minus(X''', 0)), s(0))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(X''')), s(0)) -> QUOT(s(minus(X''', 0)), s(0))


The following usable rules for innermost can be oriented:

minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)


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

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 21
Argument Filtering and Ordering


Dependency Pair:

QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(X''')), s(s(Y'))) -> QUOT(ifMinus(le(X''', Y'), s(X'''), s(Y')), s(s(Y')))


The following usable rules for innermost can be oriented:

ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)


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

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


   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
Nar
       →DP Problem 3
Nar
           →DP Problem 19
Nar
             ...
               →DP Problem 23
Dependency Graph


Dependency Pair:


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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