I encountered the following example whilst studying Equivalence Classes from How to prove it, Velleman. :
Let $ B= \{(p, q) ∈ P × P | \text{the person p has the same birthday as the person q } \} $
Let D be the set of all possible birthdays. In other words,$ D = \{Jan. 1, Jan. 2, Jan. 3, . . . ,Dec. 30, Dec. 31\}. $
Now for each $d ∈ D$, let $P_d = \{p ∈ P | \text{the person p was born on the day d } \} $ .
Then the family $ F = \{P_d | d ∈ D\} $ is an indexed family of subsets of P.
We can then redefine B as:
$B = \{(p, q) ∈ P × P | ∃d ∈ D(p ∈ P_d \land q ∈ P_d )\}$
$= \{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$
$ = ∪_{d∈D}(P_d × P_d ). $
My question: Could I write $ \{(p, q) ∈ P × P | ∃d ∈ D((p, q) ∈ P_d × P_d )\}$ as $ \{(p, q) ∈ P_d × P_d | d ∈ D\}$?
I think yes ( but I want to be sure ). My reasoning is since inside the set builder notation we have $ (p,q) ∈ P × P \land (p,q) ∈ P_d × P_d $ then $ (P × P) \land (P_d × P_d) = (P_d × P_d)$ Since $ P_d × P_d \subseteq P × P $.