Term Rewriting System R:
[Y, X]
minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> 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)
GEQ(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) -> GEQ(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
AFS
       →DP Problem 3
FwdInst


Dependency Pair:

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


Rules:


minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 4
Dependency Graph
       →DP Problem 2
AFS
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Argument Filtering and Ordering
       →DP Problem 3
FwdInst


Dependency Pair:

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


Rules:


minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 3
FwdInst


Dependency Pair:


Rules:


minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Forward Instantiation Transformation


Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))


Rules:


minus(0, Y) -> 0
minus(s(X), s(Y)) -> minus(X, Y)
geq(X, 0) -> true
geq(0, s(Y)) -> false
geq(s(X), s(Y)) -> geq(X, Y)
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) -> X
if(false, X, Y) -> Y


Strategy:

innermost




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

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
no new Dependency Pairs are created.
The transformation is resulting in no new DP problems.


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