Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs. 


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


Dependency Pair:

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


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





The following dependency pair can be strictly oriented:

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


Additionally, the following rules can be oriented:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(log(x1))=  x1  
  POL(MIN(x1, x2))=  1 + x1  
  POL(min(x1, x2))=  x1  
  POL(quot(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  

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:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





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:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





The following dependency pair can be strictly oriented:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))


Additionally, the following rules can be oriented:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(QUOT(x1, x2))=  1 + x1 + x2  
  POL(0)=  0  
  POL(log(x1))=  x1  
  POL(min(x1, x2))=  x1  
  POL(quot(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 5
             ↳Dependency Graph
       →DP Problem 3
         ↳Polo


Dependency Pair:


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





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:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





The following dependency pair can be strictly oriented:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))


Additionally, the following rules can be oriented:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(log(x1))=  x1  
  POL(min(x1, x2))=  x1  
  POL(quot(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(LOG(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 6
             ↳Dependency Graph


Dependency Pair:


Rules:


min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))





Using the Dependency Graph resulted in no new DP problems.

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