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On the Wikipedia page for the Axiom of Choice the following statement is given:

$(\forall x^\sigma)(\exists y^\tau)R(x,y)\rightarrow(\exists f^{\sigma \rightarrow \tau})(\forall x^\sigma)R(x, f(x))$

Most of it seems fairly straightforward, except for the meanings of the symbols that look like 180 degree rotated 'E' and 'A'

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    $\begingroup$ Interesting question. I don't really understand how this can be considered "fairy straightforward" if the 4 quantifiers aren't known or understood. $\endgroup$ – Eric Duminil Jan 11 '20 at 13:20
  • $\begingroup$ See here & here. $\endgroup$ – J.G. Jan 11 '20 at 15:09
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    $\begingroup$ It's worth adding that their LaTeX codes are "\forall" and "\exists" respectively. $\endgroup$ – Noah Schweber Jan 12 '20 at 17:54
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The symbols are quantifiers. They bind a new variable name to the symbolic logic statements. ∃ reads as there exists. ∀ reads for all so the first part of the statement would be read as:

forall x (of type 𝜎), there exists a y (of type 𝜏) such that ...

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    $\begingroup$ As someone who thought OP's quote looks like caveman writing, could you expand this answer to explain all the hieroglyphs? $\endgroup$ – user1717828 Jan 11 '20 at 1:47
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    $\begingroup$ @user1717828 I'm pretty sure the OP couldn't understand anything too given that they did not know the most basic elements of the formula (which they call "equation" even though there is no equal sign anywhere...). $\endgroup$ – Bakuriu Jan 11 '20 at 8:38
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    $\begingroup$ @user1717828 I'll move to Roman letters: "Let $R$ be a relation of things in $X$ to things in $Y$. If it is the case that for all $x$ in $X$ there is a $y$ in $Y$ which is related to $x$, then there is a function $X \to Y$ which takes any $x$ to a thing which is related to $x$." A relation of things in $X$ to things in $Y$ is viewed as a predicate on $X \times Y$, which answers "True" to the input $(x, y)$ if and only if $x$ is related to $y$; hence the use of the function notation $R(x, f(x))$ which means "$x$ is related to $f(x)$". $\endgroup$ – Patrick Stevens Jan 11 '20 at 8:41
  • $\begingroup$ @Bakuriu indeed. I also had the same mistake. fixed it now. $\endgroup$ – Apoorv Jan 11 '20 at 9:10
  • $\begingroup$ @Bakuriu I figured out how to interpret most of it, I just don't have the background knowledge to know that 'equation' isn't the right term. $\endgroup$ – 0x777C Jan 12 '20 at 16:49
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$\forall$ reads as "for all", and $\exists$ reads as "there exists". So, in english we have

$$\text{"if }\underbrace{\text{for all $x$}}_{\forall x^\sigma}\text{ }\underbrace{\text{exists a $y$}}_{\exists y^\tau}\text{ with $R(x,y)$, }\underbrace{\text{then}}_\to\text{ }\underbrace{\text{there is a function $f$}}_{\exists f^{\sigma\to\tau}}\text{ so that }\underbrace{\text{for all $x$}}_{\forall x^\sigma}\text{ holds $R(x,f(x))$".}$$

I skipped over the $\sigma$ and $\tau$ superscripts, as they indicate types and are not of primary importance here.

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