Term Rewriting System R:
[x]
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

HALF(s(s(x))) -> S(half(x))
HALF(s(s(x))) -> HALF(x)
S(log(0)) -> S(0)
LOG(s(x)) -> S(log(half(s(x))))
LOG(s(x)) -> LOG(half(s(x)))
LOG(s(x)) -> HALF(s(x))

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





The following dependency pair can be strictly oriented:

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


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

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Narrowing Transformation


Dependency Pair:

LOG(s(x)) -> LOG(half(s(x)))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

LOG(s(x)) -> LOG(half(s(x)))
three new Dependency Pairs are created:

LOG(s(0)) -> LOG(0)
LOG(s(s(x''))) -> LOG(s(half(x'')))
LOG(s(log(0))) -> LOG(half(s(0)))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

LOG(s(log(0))) -> LOG(half(s(0)))
LOG(s(s(x''))) -> LOG(s(half(x'')))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





The following dependency pair can be strictly oriented:

LOG(s(log(0))) -> LOG(half(s(0)))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

s(log(0)) -> s(0)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(log(x1))=  1  
  POL(s(x1))=  x1  
  POL(half(x1))=  0  
  POL(LOG(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pair:

LOG(s(s(x''))) -> LOG(s(half(x'')))


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





The following dependency pair can be strictly oriented:

LOG(s(s(x''))) -> LOG(s(half(x'')))


Additionally, the following usable rules w.r.t. to the implicit AFS can be oriented:

s(log(0)) -> s(0)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(log(x1))=  0  
  POL(s(x1))=  1 + x1  
  POL(half(x1))=  x1  
  POL(LOG(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Polo
             ...
               →DP Problem 6
                 ↳Dependency Graph


Dependency Pair:


Rules:


half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
s(log(0)) -> s(0)
log(s(x)) -> s(log(half(s(x))))





Using the Dependency Graph resulted in no new DP problems.

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