In the context of Upper bounds computaion and Big Oh Notation, I was wondering if the following could be proved... if they are equivalent.
$\mathcal{O}((log(n))^{-1}) = (\mathcal{O}(log(n)))^{-1}$
$\mathcal{O}((log(n))^{-1})$ can be rewritten as
$\mathcal{O}(1/(log(n)))$ which is
$\mathcal{O}(1)/ \mathcal{O}(log(n))$ . This can be simplified to
$1/ \mathcal{O}(log(n))$
which is the same as the RHS. Would this be a possible solution? Any mistakes that are pointed are welcome.
This kind of thing just doesn't work. For example, one of your intermediate terms is $O(1)/O(\log n)$. However any function $f$ can be written as $g/h$ where $g=O(1)$ and $h=O(\log n)$. If $f=O(1)$, then let $g=f$ and $h=1$. If $f=\Omega(1)$, then let $g=1$ and $h=1/f$.
For a more formal treatment, try to rewrite each of your statements that involve big-$O$ in terms of "There are constants $k$ and $n_0$ such that, for all $n\geq n_0$, $f(n)\leq k\,g(n)$." You'll probably find that it's not even clear what these formal statements should be and that, if you manage to figure that out, then some of the formal statements will be false.
The underlying point you've overlooked is that big-$O$ provides upper bounds but, if you have a quotient $a/b$ then upper-bounding that quotient requires upper-bounding $a$ and lower-bounding $b$. Big-$O$ can't express that lower bound.
