Exponentials of truth values

Question

I noticed that the exponentiation identity $$\exp(r + s) = \exp(r) \cdot \exp(s)~,$$ which is of course completely standard for real or complex numbers also holds in a Boolean setting.

That is, when I associate

• addition ($+$) with disjunction ($\vee$) (as is standard),

• multiplication ($\:\cdot\:$) with conjunction ($\wedge$) (as is also standard),

• and exponentiation ( $\exp(\,-\,)$ ) with negation ( $\neg$ ),

then the exponentiation identity becomes De'Morgans law: $$\neg(r \vee s) \equiv \neg r \wedge \neg s$$

I'm pretty sure this holds in Heyting algebras as well. Is this very well known? Is there something deeper behind this analogy? Or is it completely obvious? Is negation the exponentiation of logic? Or is the above a mere coincidence with no deeper meaning?

Answer 1

Yes, $(r \lor s) \implies x$ and $(r \implies x) \land (s \implies x)$ are the same truth value for any truth value $x$. In fact, using different logical notation that looks more like arithmetic, we could write this as $x^{r+s} = x^rx^s$. Thus $\implies x$ is kind of like an exponential function. The DeMorgan law you gave is the special case $x = \bot$.

They are equal because they are logically equivalent in intuitionistic logic. The fact that the second value implies the first value is known as proof by cases.

Answer 2

Is there something deeper behind this analogy?

If you have a bijection $\beta\colon X\to Y$, then to every operation on $X$ there is a ($\beta$-)conjugate operation on $Y$. Namely, if $f\colon X^k\to X$ is an operation on $X$, then ${}^{\beta}f:=\beta\circ f\circ \beta^{-1}$ is the conjugate operation on $Y$. If $\mathbb X=\langle X; f_1,f_2,\ldots\rangle$ is an algebraic structure and $\mathbb Y=\langle Y; f_1,f_2,\ldots\rangle$ is the structure of the same signature where the operations are interpreted so that $f_i^{\mathbb Y} = {}^{\beta}(f_i^{\mathbb X})$, then $\beta\colon \mathbb X\to \mathbb Y$ is an isomorphism. Every isomorphism $\beta\colon \mathbb X\to \mathbb Y$ can be viewed as a statement that the interpretation of the operations on the codomain structure are the $\beta$-conjugates of the interpretations of the operations on the domain structure.

The identity $\textit{exp}(x+y)=\textit{exp}(x)\cdot \textit{exp}(y)$ communicates that the bijection $\textit{exp}\colon \mathbb R\to \mathbb R^+$ is an isomorphism from the structure $\langle \mathbb R; +\rangle$ to the structure $\langle \mathbb R^+;\cdot\rangle$. This bijection realizes multiplication on $\mathbb R^+$ as the operation that is ($\textit{exp}$-)conjugate to addition on $\mathbb R$. Similarly, the negation operation on a Boolean algebra $B$ is a bijection $\neg\colon B\to B$ that realizes meet as the conjugate of join, join as the conjugate of meet, negation as its own conjugate, etc.

I would say that your observation is this: bijections lead to relations expressing conjugacy. That is, the assertion that ''$\textit{exp}\colon \mathbb R\to \mathbb R^+$ is a bijection and $\textit{exp}(x+y)=\textit{exp}(x)\cdot \textit{exp}(y)$ holds'' contains the same information as the statement that ''$\textit{exp}\colon \mathbb R\to \mathbb R^+$ is a bijection and multiplication on $\mathbb R^+$ is the $\textit{exp}$-conjugate of addition on $\mathbb R$''. Both of these are saying only that ''$\textit{exp}\colon \langle \mathbb R; +\rangle\to\langle \mathbb R^+;\cdot\rangle$ is an isomorphism''.

Answer 3

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.