Interpreting $\{x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\forall y~(y\in\{2,\dots,x/2\})\implies(x\not\equiv0\pmod y))\}$

Question

I am learning universal quantifiers and existential quantifiers and I am still at the beginner level. I found a question in my lecture slides as following

$$\left\{\;x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\;\forall y~[(y\in \{2,\dots,x/2\})\implies(x\not\equiv0\pmod y)]\;)\;\right\}$$

and the answer was given as

$$\{1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}$$

$\Bbb N_1$ means NAT1 which means natural numbers which are starting from $1$ ($1$ inclusive).

Unfortunately, I don't understand how does that answer come. So any help with a step by step explanation is greatly appreciated.

Answer 1

Let us decompose the whole expression step by step. First, we have our intial declaration of what type of things we are even looking for to collect in our set

$$\{x~|~x\in\Bbb N_1\}=\{1,2,3,4,\dots\}\tag1$$

So we are looking for natural numbers $x$(starting by $1$). But they have to met some other conditions too, denoted by conjunction $\land$. Second, we have a restriction concerning their size

$$\{x~|~x\in\Bbb N_1~\land~x\leq50\}=\{1,2,3,4,\dots,49,50\}\tag2$$

Now, a new variable comes into play: $y$. In contrast to $x$ its not clearly mentioned that we also have $y\in\Bbb N_1$ as all of our $y$ are defined to belong to the list $2,3,\dots,x/2-1,x/2$, or to put it in a slightly different way: we have $y\in\{2,3,\dots,x/2-1,x/2\}$. We demand that for every $y\in\{2,3,\dots,x/2-1,x/2\}$ it holds that $x\not\equiv0\mod y$ simultaneously. This is encoded by the formula

$$\forall y((y\in\{2,3,\dots,x/2-1,x/2\})\implies(x\not\equiv0\mod y))\tag3$$

The role $\forall y$ plays here is to ensure the aforementioned simultaneity. Translating this into more familiar notions: we want to have that our number $x\in\Bbb N_1$ which is not divisble by any of the preceding numbers up to $x/2$ (we exclude all $y$ such that $x/2\leq y\leq x$ as they trivially fulfill this condition); that are precisely the primes (and $1$)!

Now, take a look at our complete set description

$$M=\{x~|~x\in\Bbb N_1~\land~x\leq50~\land~(\forall y~(y\in \{2,\dots,x/2\})\implies(x\not\equiv0\pmod y))\}$$

We conclude that by $(1)$ and $(2)$ that we are looking for natural numbers $x\leq50$ and by $(3)$ such that they are not divisble by any of the preceding numbers. These are precisely the primes $\leq50$ expanded by $1$. So our solution set is

$$\therefore~M=\{1,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\}$$