Term Rewriting System R:
[x, y]
sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

SQR(s(x)) -> +'(sqr(x), s(double(x)))
SQR(s(x)) -> SQR(x)
SQR(s(x)) -> DOUBLE(x)
SQR(s(x)) -> +'(sqr(x), double(x))
DOUBLE(s(x)) -> DOUBLE(x)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:

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


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

+'(x, s(y)) -> +'(x, 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(s(x1))=  1 + x1  
  POL(+'(x1, x2))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 4
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polynomial Ordering
       →DP Problem 3
         ↳Polo


Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

DOUBLE(s(x)) -> DOUBLE(x)


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(DOUBLE(x1))=  x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polynomial Ordering


Dependency Pair:

SQR(s(x)) -> SQR(x)


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

SQR(s(x)) -> SQR(x)


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(s(x1))=  1 + x1  
  POL(SQR(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph


Dependency Pair:


Rules:


sqr(0) -> 0
sqr(s(x)) -> +(sqr(x), s(double(x)))
sqr(s(x)) -> s(+(sqr(x), double(x)))
double(0) -> 0
double(s(x)) -> s(s(double(x)))
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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