Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
LT(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> IF(lt(x, y), 0, s(div(-(x, y), s(y))))
DIV(s(x), s(y)) -> LT(x, y)
DIV(s(x), s(y)) -> DIV(-(x, y), s(y))
DIV(s(x), s(y)) -> -'(x, y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 4
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. -'(s(x), s(y)) -> -'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 5
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. LT(s(x), s(y)) -> LT(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(0, s(y)) -> 0
-(s(x), s(y)) -> -(x, y)
lt(x, 0) -> false
lt(0, s(y)) -> true
lt(s(x), s(y)) -> lt(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
div(x, 0) -> 0
div(0, y) -> 0
div(s(x), s(y)) -> if(lt(x, y), 0, s(div(-(x, y), s(y))))


Strategy:

innermost




As we are in the innermost case, we can delete all 8 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 6
             ↳Negative Polynomial Order


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following Dependency Pair can be strictly oriented using the given order.

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


Moreover, the following usable rules (regarding the implicit AFS) are oriented.

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


Used ordering:
Polynomial Order with Interpretation:

POL( DIV(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( -(x1, x2) ) = x1

POL( 0 ) = 0


This results in one new DP problem. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 6
             ↳Neg POLO
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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