Term Rewriting System R:
[x, y]
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MIN(s(x), s(y)) -> MIN(x, y)
MAX(s(x), s(y)) -> MAX(x, y)
-'(s(x), s(y)) -> -'(x, y)
GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) -> -'(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) -> MAX(x, y)
GCD(s(x), s(y)) -> MIN(x, y)

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 5
             ↳Dependency Graph
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





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
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
           →DP Problem 6
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





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
       →DP Problem 4
         ↳Remaining


Dependency Pair:

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


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the implicit AFS that need to be oriented.

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

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
           →DP Problem 7
             ↳Dependency Graph
       →DP Problem 4
         ↳Remaining


Dependency Pair:


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Polo
       →DP Problem 3
         ↳Polo
       →DP Problem 4
         ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))


Rules:


min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))




Termination of R could not be shown.
Duration:
0:00 minutes