In this context
${n \in \mathbb{N}| (x > 1) \wedge (\forall x,y \in \mathbb{N})[(xy= n) \Rightarrow (x=1) \lor (y=1)]}$
what is the meaning of the right part of the first logical conjunction: ${(\forall x,y \in \mathbb{N})[(xy= n) \Rightarrow (x=1) \lor (y=1)]}$? More specifically, what does the square brackets mean, and also the right arrow?
Thanks
$\endgroup$ 13 Answers
$\begingroup$The right arrow is for "implies". The brackets are just used in place of parentheses for clarity.
$\endgroup$ $\begingroup$I guess you are trying to describe the set of prime numbers. This is set-builder notation, but it needs surrounding braces for that to be clear. As Daniel Fischer indicates in the comments, the square brackets are the same as parentheses and used for visual distiguishability, although all of the parentheses and brackets in the expression are redundant considering rules of precedence. It doesn't make sense to refer to $x$ outside the scope of the $\forall x$ quantifier, and if you are trying to describe the primes then that clause should be $n \gt 1$ instead of $x \lt 1$ in order to exclude $1$ as a prime. So another way to say what I guess you mean is:
$\{n \in \mathbb{N} | n \gt 1 ∧ \forall x, y \in \mathbb{N}: x \cdot y = n \implies x = 1 ∨ y = 1 \}$
which is an expression for the set of primes.
$\endgroup$ 2 $\begingroup$Square brackets also denote equivalence classes in set theory:
$\endgroup$Given a set $S$ and an equivalence relation $\sim$ on $S$, theequivalence class of an element $a$ in $S$, denoted by $\textbf{[a]}$, is the set $\{ x ∈ S \space | \space x \sim a \}$
