Is this a valid use of big-O notation?

Question

Suppose that $m=O(n^{c+1/2})$ for some real $c>0$ and $x=O(\sqrt{\log m})$. Are the following two computations valid? I understand that I'm abusing notations a bit to get at the desired results.

Computation 1 \begin{align} f(x)\leq \exp(-x^2/2)&=\exp(-O(\log m))\\ &=\exp(-O(\log O(n^{c+1/2})))\\ &=\exp(-O(\log n^{c+1/2}))\\ &\leq \exp(-K_1\log n^{c+1/2})))\\ &=n^{-O(1)} \end{align} where $K_1>0$ when $n$ is sufficiently large (i.e. there exists $K_1>0$, $n_0$ such that when $n\geq n_0$, $\exp(-O(\log n^{c+1/2}))\leq \exp(-K_1\log n^{c+1/2})))$. The last quantity $n^{-O(1)}$ represents the set of polynomially-bounded functions (see accepted answer in this post).

Computation 2 $$ f(x)\frac{m}{n^{1/2}} \leq n^{-O(1)}\frac{m}{n^{1/2}}=\leq n^{-O(1)}\frac{m}{n^{c+1/2-c}}=O(1)?? $$

Answer 1

If these computations appear in a paper for me to review, I would like to ask the author to rewrite.

$$\exp(-x^2/2)\leq \exp\left(-K_1\log n^{c+1/2\,}\right), \text{ where } K_1>0$$

This is wrong. The direction of inequality should be switched. For example, let $n\to\infty$ and $x$ be constant 1, the left hand side remains a constant while the right hand side goes to 0. This error means "Computation 2" is wrong from the start.

The last quantity $n^{-O(1)}$ represents the set of polynomially-bounded functions.

Instead of $n^{-O(1)}\,$, $n^{O(1)}$ represents asymptotically the set of polynomially-bounded functions.