Analyzing logical form clarification

Question

I want to refer to Example 2.3.1 of Velleman's book "How to prove it" . It is asked to analyze the logical form of $\{x_i\; | \; i\in I\} \subseteq A$. Two possible answers are given. The first one is $\forall x(\exists i\in I(x=x_i) \rightarrow x\in A)$, which I understand. The second one is $\forall i\in I(x_i\in A)$. I don't understand how to go from the first to the second. I would appreciate any guide. Thank you in advance.

Answer 1

The two statements are equivalent. To understand why, it is better to have a closer look at the logical form of the two statements.

First, a clarification. A statement of the form $\forall i \!\in\! I \, P(i)$ is just a shorthand for $\forall i (i \in I \to P(i))$. Similarly, a statement of the form $\exists i \!\in\! I \, P(i)$ is just a shorthand for $\exists i (i \in I \land P(i))$.

Keeping in mind this, let us see how to prove that the first statement

$\forall x(\exists i \in I(x=x_i) \to x \in A)$, that is, $\forall x (\exists i (i \in I \land x =x_i) \to x \in A)$),

is equivalent to the second statement

$\forall i \in I (x_i \in A)$, that is, $\forall i (i \in I \to x_i \in A))$,

using general laws of logical equivalences.

\begin{align} &\quad \ \forall x (\exists i (i \in I \land x =x_i) \to x \in A) & \\ & \equiv \forall x \forall i ((i \in I \land x = x_i) \to x \in A) &&\text{(because $(\exists i P(i)) \to Q \equiv \forall i (P(i) \to Q)$)} \\ & \equiv \forall x \forall i (i \in I \to (x = x_i \to x \in A)) &&\text{(because $(P \land Q) \to R \equiv P \to (Q \to R)$)} \\ &\equiv \forall i (i \in I \to \forall x (x = x_i \to x \in A)) &&\text{(because $\forall x(P \to Q(x)) \equiv P \to \forall xQ(x)$)} \\ &\equiv \forall i (i \in I \to x_i \in A)) &&\text{(because of the laws for $=$)} \end{align}