How can you formularize "if"-conditions in a language-agnostic way?

I'm talking about formulas like

$$ x=\begin{cases} 2,&\text{ if } n \le 2 \lor m=3\\ 3,&\text{ if } n \gt 2 \land m=n \end{cases} $$ Of course, everybody in the world will understand the "if", but is there a way to write it completely without using any "real world" language whatsoever?

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    $\begingroup$ There's no need to use such a word, at least not in your example, AFAICT. $\endgroup$
    – Gigili
    May 7, 2012 at 8:59
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    $\begingroup$ It is a common mistake to think that mathematics is written in pure symbols. There is a lot of text around and it just makes things clearer. $\endgroup$
    – Asaf Karagila
    May 7, 2012 at 9:01
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    $\begingroup$ If you're going to use "if", you may as well use "and" and "or", and I assure you it looks much better that way. $\endgroup$
    – Zhen Lin
    May 7, 2012 at 9:01
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    $\begingroup$ To add on to Asaf's point, symbols are used because sometimes they make things clearer to express, not because they are inherently better than words. For example, it's easier to understand "$y = ax + b$" than to understand "$y$ is the sum of $b$ and the product of $a$ and $x$", but it is easier to understand "$A$ is a nonsingular $3\times3$ matrix" than to understand "$A \in M_{3\times3} \wedge \det A \ne 0$". $\endgroup$
    – user856
    May 7, 2012 at 9:13
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    $\begingroup$ Asaf has told you about how words are a lot of times preferred in mathematical writing (and personally I'm not at all fond of $\land$ and $\lor$); that being said, look into Iverson brackets. $\endgroup$ May 7, 2012 at 9:15

2 Answers 2


If you absolutely insist, you can use Iverson brackets:

$$x=\begin{cases} 2,&\text{ if } n \le 2 \lor m=3\\ 3,&\text{ if } n \gt 2 \land m=n \end{cases}$$

is equivalent to


But the first version is easier to read, and while I don't at all mind $\lor$ and $\land$, $\text{or}$ and $\text{and}$ are preferable for most audiences.


I was taught that if ... then is written in such form:

$$ x>0 \Rightarrow y=2x $$

which means if $x$ is greater than 0 then assign $2x$ to $y$.


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