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(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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)
LE(s(x), s(y)) -> LE(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) -> MINUS(s(x), s(y))

Furthermore, R contains three SCCs.


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


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


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


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(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(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: 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
Rw


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


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
Rw


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(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(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 2
FwdInst
           →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 3
Rw


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(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(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 2
FwdInst
           →DP Problem 7
FwdInst
             ...
               →DP Problem 8
Polynomial Ordering
       →DP Problem 3
Rw


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


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


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
Rewriting Transformation


Dependency Pair:

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

QUOT(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
Rw
           →DP Problem 10
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)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(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(x''), s(0)) -> QUOT(x'', s(0))
QUOT(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
Rw
           →DP Problem 10
Nar
             ...
               →DP Problem 11
Polynomial Ordering


Dependency Pair:

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost that need to be oriented.

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

resulting in one new DP problem.



   R
DPs
       →DP Problem 1
FwdInst
       →DP Problem 2
FwdInst
       →DP Problem 3
Rw
           →DP Problem 10
Nar
             ...
               →DP Problem 13
Dependency Graph


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


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
Rw
           →DP Problem 10
Nar
             ...
               →DP Problem 12
Polynomial Ordering


Dependency Pair:

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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


Additionally, the following usable rules for innermost 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)=  1  
  POL(minus(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.


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