Uppercase E notation for sets?

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In Jónsson and Tarski's (1951) paper Boolean Algebras with Operators, Part I from the American Journal of Mathematics, they write formulae such as

$L_i = \underset{u}{\mathbf{E}} \, [u \in At^m \text{ and } u \leq x^{(i)}]$

and

$K = \underset{u}{\mathbf{E}} \, [y \geq u \in At^m]$,

without explaining this $\underset{u}{\mathbf{E}}$ notation. From the context, I guess these define sets, i.e., they respectively mean

$L_i = \{u \mid u \in At^m \text{ and } u \leq x^{(i)} \}$

and

$K = \{u \mid y \geq u \in At^m \}$.

Am I correct in this?

Also, is this notation something common that mathematicians generally understand? What does E stand for, and where did this notation originate? Are there good books/articles/webpages where I can learn about this notation?

I would greatly appreciate your help!

Additional note. To give some context, on p. 900, following the first formula, they proceed to define $K = \bigcup_{i \in I} L_i$ (which has nothing to do with the $K$ in the second of the formulae above) and say that $u \in K$ if and only if $\sum_{i \in I} x^{(i)} \geq u \in At^m$. In order for this equivalence to hold, it seems to me that $L_i = \{u \mid u \in At^m \text{ and } u \leq x^{(i)} \}$.

$At$ denotes the set consisting of $0$ and all the atoms of a Boolean algebra $A$. For my interpretation to make sense, however, I suppose that $At^m$ should be the set consisting of $0$ and all the atoms of $A^m$, and not the $m$-times product of $At$.

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This is a rather old paper, so it is not a surprise to find some outdated notation. AFAIK, this $\mathbf{E}$ is not in frequent use anymore, may mean something like ensemble (French for “set”).

Indeed, we have the equality $$ \underset{u}{\mathbf{E}} \, [u \in At^m \text{ and } u \leq x^{(i)}] = \{u \mid u \in At^m \text{ and } u \leq x^{(i)}\}, $$ but you have to take into account that each $x^{(i)}$ is an $m$-tuple of elements of $A$, and the notation $u \leq x^{(i)}$ is defined by the “pointwise” comparison, that is, $u_j\leq x^{(i)}_j$ for all $j$. Then $At^m$ is actually an $m$-tuple of atoms.