A way to formalize what you are noticing is the following:

So let $Q$ be a group with a single generator. Let $X$ be some set of your choice. We can then define a "blk structure" as follows:

A blk structure is a set $X$ and group $Q$ along with a set of binary associative operators $B_q: X^2 \rightarrow X $ indexed by the elements $Q$ with a transition map $T$ s.t. $T(B_q(u,v)) = B_{q+1}(T(u),T(v))$

Over the reals we have $X=\mathbb{R}$, $Q=\mathbb{Z}$, $B_q(x,y) = \text{exp}^{q}(\text{exp}^{-q}(x) + \text{exp}^{-q}(y))$, $T=\text{exp}$ (Note here $\text{exp}^q$ means the $q^{\text{th}}$ iterate of exponentiation).

To clarify then we have the following:
$$\begin{array}{l} \vdots \\ B_{-1}(x,y) = \ln(e^{x}+e^{y}) \\ B_0(x,y) = x+y \\ B_1(x,y) = xy \\ B_2(x,y) = e^{\ln(x)\ln(y)} \\ \vdots \end{array} $$

And it's clear that $e^{B_q(x,y)} = B_{q+1}(e^x, e^y)$.

In formal logic we have $X = \lbrace0, 1 \rbrace $, $Q=\mathbb{Z}_2$, $B_0 = \vee, B_1 = \wedge$, and $T=\neg$. Spelled out we then have

$$ \begin{array}{l} B_0(x,y) = x \vee y \\ B_1(x,y) x \wedge y \end{array} $$

And similarly $\neg B_a(x,y) = B_{a + 1} (\neg x, \neg y)$

I'm not sure about this next part but I would hope that just like we can quotient the integers by the evens to get $\mathbb{Z}/2\mathbb{Z} = \mathbb{Z}_2$ that there might exist generally a "natural" way to quotient the $\mathbb{R}$-blk structure by some suitable relation to get the $\lbrace 0, 1 \rbrace$-blk structure of formal logic. And if you can find that construction and its "natural" you would have substantially clarified the matter.

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