Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

HALF(s(s(x))) -> HALF(x)
LASTBIT(s(s(x))) -> LASTBIT(x)
CONV(s(x)) -> CONV(half(s(x)))
CONV(s(x)) -> HALF(s(x))
CONV(s(x)) -> LASTBIT(s(x))

Furthermore, R contains three SCCs. 


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


Dependency Pair:

HALF(s(s(x))) -> HALF(x)


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))





The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(x)


Additionally, the following rules can be oriented:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(conv(x1))=  0  
  POL(0)=  0  
  POL(HALF(x1))=  1 + x1  
  POL(cons(x1, x2))=  0  
  POL(lastbit(x1))=  1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(half(x1))=  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
         ↳Remaining


Dependency Pair:


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

LASTBIT(s(s(x))) -> LASTBIT(x)


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))





The following dependency pair can be strictly oriented:

LASTBIT(s(s(x))) -> LASTBIT(x)


Additionally, the following rules can be oriented:

half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(conv(x1))=  0  
  POL(0)=  0  
  POL(cons(x1, x2))=  0  
  POL(lastbit(x1))=  1  
  POL(nil)=  0  
  POL(s(x1))=  1 + x1  
  POL(half(x1))=  x1  
  POL(LASTBIT(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
         ↳Remaining


Dependency Pair:


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))





Using the Dependency Graph resulted in no new DP problems.


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




The following remains to be proven:
Dependency Pair:

CONV(s(x)) -> CONV(half(s(x)))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
lastbit(0) -> 0
lastbit(s(0)) -> s(0)
lastbit(s(s(x))) -> lastbit(x)
conv(0) -> cons(nil, 0)
conv(s(x)) -> cons(conv(half(s(x))), lastbit(s(x)))




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