Logic formula for exactly n unique objects (no more, no less)

Question

I have a question in Logic:

If I am asked to construct a formula, using the '=' predicate, that shows that there are exactly n objects, I need to show that there are no n+1 objects, right?

For example, to show that there are exactly 3 objects, I will show that there are no 4 objects.

Also, I need to show that there exist n objects.

My question is, do I need to show that there is no case where only n-1 objects exist?

For example, to show that there are exactly 3 objects, do I need to show that there is no case in which there are only 2 objects?

If so, is that the correct form to do so? :

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1) Showing that there is no case of 4 elements:

$ {\lnot}{\exists}x{\exists}y{\exists}z{\exists}w(x{\neq}y{\land}x{\neq}z{\land}x{\neq}w{\land}y{\neq}z{\land}y{\neq}w{\land}z{\neq}w) $

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2) Showing that there are 3 elements:

$ {\exists}x{\exists}y{\exists}z(x{\neq}y{\land}x{\neq}z{\land}y{\neq}z) $

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3) Showing that there is no case of only 2 elements:

$ {\lnot}{\exists}x{\exists}y{\lnot}{\exists}z(x{\neq}y{\land}x{\neq}z{\land}y{\neq}z) $

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And finally, combining the three:

$ (1){\land}(2){\land}(3) $

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I am really not sure.

Thanks in advance

EDIT:

Actually, assuming that I do need to show that there is no case of less than 3 elements (n elements), I would probably need to show that there is no case where there is just one element as well (from 1 until n-1), am I correct?

Answer 1

You need to say that (a) there are at least $n$ elements, and (b) there are at most $n$ elements. To express (a), $$ L_n := \exists x_1\dotsc \exists x_n\, \left( \bigwedge_{1\le i < j \le n} x_i\ne x_j \right). $$ To express (b), $$ M_n := \forall x_1\dotsc \forall x_{n+1}\, \left( \bigvee_{1\le i < j \le n+1}x_i = x_j\right). $$ So the sentence $L_n \land M_n$ holds iff there are exactly $n$ elements.