Term Rewriting System R:
[x, y]
top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TOP(sent(x)) -> TOP(check(rest(x)))
TOP(sent(x)) -> CHECK(rest(x))
TOP(sent(x)) -> REST(x)
CHECK(sent(x)) -> CHECK(x)
CHECK(rest(x)) -> REST(check(x))
CHECK(rest(x)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(cons(x, y)) -> CHECK(y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Remaining


Dependency Pairs:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)
CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)





The following dependency pairs can be strictly oriented:

CHECK(cons(x, y)) -> CHECK(y)
CHECK(cons(x, y)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(sent(x)) -> top(check(rest(x)))
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  1  
  POL(rest(x1))=  x1  
  POL(cons(x1, x2))=  1 + x1 + x2  
  POL(check(x1))=  x1  
  POL(nil)=  0  
  POL(CHECK(x1))=  x1  
  POL(sent(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pairs:

CHECK(rest(x)) -> CHECK(x)
CHECK(sent(x)) -> CHECK(x)


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)





The following dependency pair can be strictly oriented:

CHECK(rest(x)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(sent(x)) -> top(check(rest(x)))
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(rest(x1))=  1 + x1  
  POL(cons(x1, x2))=  x2  
  POL(check(x1))=  x1  
  POL(nil)=  0  
  POL(CHECK(x1))=  x1  
  POL(sent(x1))=  x1  

resulting in one new DP problem. 



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


Dependency Pair:

CHECK(sent(x)) -> CHECK(x)


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)





The following dependency pair can be strictly oriented:

CHECK(sent(x)) -> CHECK(x)


Additionally, the following rules can be oriented:

top(sent(x)) -> top(check(rest(x)))
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(top(x1))=  0  
  POL(rest(x1))=  1 + x1  
  POL(cons(x1, x2))=  x2  
  POL(check(x1))=  x1  
  POL(nil)=  0  
  POL(CHECK(x1))=  x1  
  POL(sent(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)





Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pair:

TOP(sent(x)) -> TOP(check(rest(x)))


Rules:


top(sent(x)) -> top(check(rest(x)))
rest(nil) -> sent(nil)
rest(cons(x, y)) -> sent(y)
check(sent(x)) -> sent(check(x))
check(rest(x)) -> rest(check(x))
check(cons(x, y)) -> cons(check(x), y)
check(cons(x, y)) -> cons(x, check(y))
check(cons(x, y)) -> cons(x, y)




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