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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

HALF(s(s(x))) -> HALF(x)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost 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
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Narrowing Transformation


Dependency Pair:

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


Rules:


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


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 7
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


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


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


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

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


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  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
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 8
                 ↳Dependency Graph


Dependency Pair:


Rules:


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


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

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