Determining whether 2 logical statements are equivalent I recently encountered a problem which is to determine whether the following statements are logically equivalent to each other. The statements are:
$$\text{$(\exists x\in U)[P(x)\land Q(x)]$ and $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$}$$
Suppose $(\exists x\in U)[P(x)\land Q(x)]$ is true. Let $x = a \in U$ s.t. $P(a)$ and $Q(a)$ is true. So both $P(a)$ and $Q(a)$ are true.
Suppose $(\exists x\in U)[P(x)\land Q(x)]$ is false. Then its negation: $(\forall x \in U)[\neg P(x) \lor\neg Q(x)]$ is true. The negation holds true whenever $P(x)$ is false or $Q(x)$ is false for any $x \in U$.
Let $x \in U$ s.t. $\neg P(x) \lor\neg Q(x)$ is true, hence, we consider 2 cases:
Case 1: $P(x)$ is false
Since $P(x)$ is false for any $x\in U$, $(\exists x\in U)P(x)$ is false, hence $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$ must be false.
Case 2: $Q(x)$ is false
Since $Q(x)$ is false for any $x\in U$, $(\exists x\in U)Q(x)$ is false, hence, $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$ must be false.
Since when $(\exists x\in U)[P(x)\land Q(x)]$ is true, $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$ is also true. And when $(\exists x\in U)[P(x)\land Q(x)]$ is false, $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$ is also false.
Therefore we can conclude that they are logically equivalent.
I would like to know whether my answer is valid and whether there is bug in my answer. Thanks!!
 A: The two statements are not equivalent, since if

*

*$Px:=x$ is even

*$Qx:=x$ is odd

*$U:=$ the set of positive integers

then the left statement is false but the right statement true.

I would present your argument more readabily by rewriting many of the symbols (for example, the quantifiers) as words, and perhaps using fewer parentheses (illustrated below). The human (as opposed to machine) reader will appreciate this.
Anyway, the first part of your proof is correct, but in the second part you have actually— sort of —proven $$∀x{\in}U\;\Big((¬Px ∨ ¬Qx)   →       ¬(Px  ∧  Qx)     \Big) \tag1 $$ instead of the (false) statement $$∀x{\in}U\;(¬Px ∨ ¬Qx)   →       ¬\Big((   ∃x{\in}U\;Px)  ∧    (∃x{\in}U\;Qx)     \Big)  .$$

Let $x \in U$ s.t. $\neg P(x) \lor\neg Q(x)$ is true

Here, you are considering one arbitrary element of $U;$ let's call it $a.$

, hence, we
consider 2 cases:
Case 1: $P(x)$ is false
Since $P(x)$ is false for any $x\in U$,

Here, you should still be referring to the specific element $a.$

$(\exists x\in U)P(x)$ is
false, hence $[(\exists x\in U)P(x)]\land[(\exists x\in U)Q(x)]$ must
be false.

Logically, the part before "hence" is just P(a) is false, while the part after "hence" is just P(a)∧ Q(a) is false.
Case 2 contains the same error.
Then, at the end, since $a$ is an arbitrary element of $U,$ we can correctly conclude that $(1)$ is true. This entire proof is irrelevant to disproving the given statement, but I'm critiquing it to reveal the "bug" (heheh) that you are asking about.

The negation holds true whenever $P(x)$ is false or $Q(x)$ is false for any $x \in U$.

I had to read this twice because it is written like it could mean either of these:
$$(\forall x\;Sx)\implies T\\
\forall x\;(Sx\implies T)$$
A: Let $A$ be the set $\{x|P(x)\}$ and $B$ be the set $\{x|Q(x)\}$
The first premise is therefore $A\cap B\ne\emptyset$ and the second $A\ne\emptyset\land B\ne\emptyset$.
The second premise doesn't enforce the first (if $x\in A$ then $x\in B$ is indeterminate).
For example $A=\{1\}, B=\{2\}$.
