I have written several articles about a way to do classical calculus without using variables, e.g. how to differentiate or integrate functions. I was asked in the comments how to do quantification without variables—you know, expressions like $∀x > 0$ $∃y > 0: y^2 = x$.

It may seem impossible to do that without using variables, but it turns out it can be done in **two different ways**, one of which closely resembles the traditional approach while the other looks somewhat strange to an unaccustomed eye.

## The “strange” way

Let’s start with the strange one, so that you get used to working with images of mappings that we will also need later. Let $A$ be a set and $f$ a function. We will use the following notation:

$$ f(A) = \{f(x): x ∈ A\}\,. $$That is, $f(A)$ is the image of $A$ under the mapping $f$. The universal quantifier can be translated as follows:

$$ ∀x ∈ A: f(x) ∈ B \quad⇔\quad f(A) ⊆ B\,. $$In particular, we can write

$$ ∀x ∈ A: f(x) = c \quad⇔\quad f(A) = \{c\}\,. $$Pretty much any universal quantification you can imagine can be written in this way. The existential quantifier is similar:

$$ ∃x ∈ A: f(x) ∈ B \quad⇔\quad B ∩ f(A) ≠ ∅\,, $$in particular

$$ ∃x ∈ A: f(x) = c \quad⇔\quad c ∈ f(A)\,. $$How do we combine the two? Suppose you had to rewrite $∀x > 0\,∃y > 0: y^2 = x$ without variables—how would you go about it? You can simply do it in two steps! First, rewrite the inner quantifier:

$$ ∀x > 0: x ∈ ((0,∞))^2 $$where $(A)^2$ means the image of $A$ under $\id^2$ (the “square function”). Now, “$∀x ∈ A: x ∈$” is the same as “$A ⊆$”, so we can rewrite the last expression as

$$ (0,∞) ⊆ ((0,∞))^2\,. $$Quite neat, isn’t it? It expresses the same idea using a very different kind of thinking. The proof would be quite different, too. Instead of manipulating algebraic expressions, we would prove that $((0,∞))^2 = (0,∞)$ using properties of $\id^2$ (which could be done just by playing around with sets, as all of its properties can be reformulated using the process just described).

## The “normal” way

Another way to look at variables is that they are just projections from a larger space. For example, when working with expressions like $x^2+y^2$, we can think of $x$ and $y$ as functions $ℝ^2 → ℝ$ defined by $x(a,b) = a$ and $y(a,b) = b$. There is a useful notation for working with such functions commonly used in probability theory. Let $R$ be relation (e.g. $<$, $>$, $=$) and $f$ and $g$ functions, then

$$ [f\,R\,g] = \{x: f(x)\,R\,g(x)\}\,, $$for example

$$ [{\sin}>0] = ⋃_{k ∈ 2ℤ} (kπ,(k+1)π)\,. $$So, if $x,y$ are the aforementioned projections, what’s

$$ [x^2 + y^2 = 1]\,? $$ The unit circle, of course. Say, we wanted to write down the fact that for each point $(a,b)$ on the unit circle, $|ab| ≤ ÷{1}{2}$. Traditionally, we would write: $$ ∀x,y ∈ ℝ, x^2 + y^2 = 1: |xy| ≤ ÷{1}{2} $$(in fact, the “correct” way of writing that would be $∀x ∈ ℝ\,$ $∀y ∈ ℝ:$ $(x^2 + y^2 = 1)$ $⇒$ $(|xy| ≤ ÷{1}{2})$, but I assume everyone understands the shortened version). How to write it without quantification? All we want to express is that the values of $|xy|$ stay within $[0,÷{1}{2}]$ (the closed interval), so we would write

$$ |xy|\big([x^2 + y^2 = 1]\big) ⊆ [0,{\textstyle ÷{1}{2}}]\,. $$If you don’t like the notation for the image of a set, you can also write:

$$ [x^2 + y^2 = 1] ⊆ \big[|xy| ≤ {\textstyle ÷{1}{2}}\big] $$which means that the unit circle lies in the region where $|xy| ≤ ÷{1}{2}$, which is logically equivalent to the previous expression. Nevertheless, there’s a semantic difference between the two; in the first case, we emphasize that the function $|xy|$ has a certain property, whereas in the second case, we emphasize that it is the set $[x^2 + y^2 = 1]$ that has a certain property.

Similarly, going back to the $∀x > 0\,∃y > 0: y^2 = x$ example, understanding the $x$ and $y$ as projections, we can write it as

$$ x\big([y > 0] ∩ [y^2 = x]\big) ⊇ (0,∞)\,. $$It looks crazy at first, but if one understands the underlying logic, it’s actually quite simple: $[y> 0] ∩ [y^2 = x]$ is the set of points $(x,y)$, $y > 0$ that are solutions to $y^2 = x$. Our assertion says that such a solution exists for each $x ∈ (0,∞)$, hence we claim that the projection of $[y > 0] ∩ [y^2 = x]$ to the $x$-axis contains $(0,∞)$.