From Velleman's 'How to Prove it' book, there is one statement - written below - of which I don't know how to write the logical form of, and I'm wondering if somebody could write it out.
The statement is:
" Prove that there is a unique $A\in\mathscr{P}(U)$ such that for every $B\in\mathscr{P}(U), A\cup B = B)$ "
I know it must be in the form $\exists{!xP(x)}$ (and of course I understand that there are different ways to cash out this statement in its full form) but I'm confused as to what represents the $P(x)$ in this case. My guess for the 'existence' part of the formula is:
$\exists{A}(A\in\mathscr{P}(U)\land \forall{B}(B\in\mathscr{P}(U) \land A\cup{B}=B))$
I know I need to do the uniqueness part as well, but I'm not even sure if this is right in the first place.
EDIT: Sorry I meant to put a conditional where the second conjunction symbol is. So:
$\exists{A}(A\in\mathscr{P}(U)\land \forall{B}(B\in\mathscr{P}(U) \to A\cup{B}=B))$
However I don't know if this is still correct for the 'existence' part of the formula, though that was what I meant to write.
Uniqueness part (in words): For all $A_1,A_2 \in \mathscr{P}(U)$, if { [for all $B$ ($B \in \mathscr{P}(U)$ implies $A_1 \cup B = A_1$] and [for all $B$ ($B \in \mathscr{P}(U)$ implies $A_2 \cup B = A_2$] } then $A_1 = A_2$.
Now to symbolize this is direct.