$$ \exists x \exists y \exists z (x\neq y \land x \neq z \land y \neq z \land P(x) \land P(y) \land P(z)) $$
I would translate the above sentence as, "There exist 3 different elements (x,y,z) for which P(x), P(y), and P(z) are true."
Does this mean at least 3 elements or exactly 3 elements? I think it's at least 3.
If so, how do I formally write exactly 3 elements?
$P$ holds for at least three elements: $$\exists x\exists y\exists z\ [x\ne y\land x\ne z\land y\ne z\land P(x)\land P(y)\land P(z)]$$ $P$ holds for at most three elements: $$\forall w\forall x\forall y\forall z\ [P(w)\land P(x)\land P(y)\land P(z)\to (w=x\lor w=y\lor w=z\lor x=y\lor x=z\lor y=z)]$$ $P$ holds for exactly three elements: $$\exists x\exists y\exists z\forall w\ [x\ne y\land x\ne z\land y\ne z\land (P(w)\leftrightarrow(w=x\lor w=y\lor w=z))]$$
$$ \exists x \exists y \exists z (x\neq y \land x \neq z \land y \neq z \land P(x) \land P(y) \land P(z)) \tag1$$
I would translate the above sentence as, "There exist 3 different elements (x,y,z) for which P(x), P(y), and P(z) are true." Does this mean at least 3 elements or exactly 3 elements?
Sentence $(1)$ has not actually identified any particular object; $x,y,z$ are just dummy variables (rather than constants or free variables).
Sentence $(1)$ means that $P(x)$ is true of three particular/distinct/different objects.
Being true of three distinct objects does not preclude being true of seven distinct objects, so does not imply being true of exactly three objects, which can be formalised as $$∃a\:∃b\:∃c\:\Big(a\ne b\,∧\, a\ne c\,∧\,b\ne c\,∧\,∀s\,\big(P(s)\,↔\,s= a\,\lor\, s=b\,\lor\, s=c\big)\Big).\tag2$$
Thus, Sentence $(1)$ means that $P(x)$ is true of at least three objects.
Incidentally, Sentence $(1)$ is logically equivalent to this more concise sentence: $$∀a\:∀b\:∃s\:\big(s\ne a\,∧\,s\ne b\,∧\,P(s)\big).\tag{1b}$$
It says "at least three elements". Imagine a structure with four elements in which $P$ is true for all of them; clearly the sentence is satisfied in this structure.
A quick way to say "precisely three" is "at least three, and not at least four".
Is there a more elegant way?
– thadeuszlay Sep 21 '16 at 22:04Something like $$\forall a\,\big(P(a)\Rightarrow (a=x)\vee (a=y)\vee (a=z)\big)$$ says that the property $P(a)$ is only satisfied if $a=x,y,z$.
x=x+1 (so, $P(x)$ is categorically false), the OP's given sentence is false whereas your suggested sentence is true.
– ryang
Jun 03 '23 at 09:08
First establish a set B to which $x,y,z$ may belong then
\begin{equation} \exists A\subset B\,(\,\vert A\vert=3 \land (\wedge_{x\in A}P(x)\,) \end{equation}
Okay, no-ones suggesting this and maybe it isn't valid but...
isn't $\exists ! a$ meaning there exists a unique $a$ acceptable, and....
isn't $\exists \{a,b,c\}$ meaning there exists a set f three elements acceptable?
If so (and maybe those notations aren't acceptable) wouldn't $\exists! \{x,y,z\} ...$ do it? There is a unique set of three elements where the conditions hold so there are three elements. If there were 4 or more elements where the conditions hold and we can make multiple different sets of 3 where the conditions hold. But as any set of three must be unique, there can only be those three elements.