Existential and Universal quantifier, what would empty sets means in combination

Question

I'm trying to understand the combination of quantifiers in a situation such as this one:

$$\exists x \in \mathbb{R} \text{ s.t. } 0 < 1, \forall y \in \mathbb{R} \text{ s.t. } 1< 0, \text{ s.t. } x \geq y$$

The first existential part is always true.
The universal part is always false. And i don't quite understand how to even process the last part now.

Once the existential is true, and the universal is false (or is an empty set), is the statement automatically true?

Answer 1

Long comment (referring also to comments above).

My suggestion is to use the rules for "unwinding" restricted quantifiers (omitting the ref to the domain: "$\in \mathbb R$" for simplicity).

Formula $∃x∈R \text { s.t. } 0<1 ∀y∈R \text { s.t. } \ldots)$ is:

$∃x \ [0<1 ∧ ∀y \ (1<0 \to x \ge y)]$.

How to negate it?

First we have to use rules for negating quantifiers, followed by propositional equivalences.

Your first step is correct:

$∀x \ [0 \ge 1 \lor ∃y \ \lnot (1<0 \to x \ge y)]$.

Now we will use: $\lnot (p \to q) \equiv (p \land \lnot q)$, to get:

$∀x \ [0 \ge 1 \lor ∃y \ (1<0 \land x < y)]$.

But this in turn is, using $(p \lor q) \equiv (\lnot p \to q)$:

$∀x \ [0 < 1 \to ∃y \ (1<0 \land x < y)]$,

and this is exactly what you expected.

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Now, for the original questions:

Once the existential is true, and the universal is false (or is an empty set), is the statement automatically true?

The answer is: Yes.

In formula $∃x \ [0<1 ∧ ∀y \ (1<0 \to x \ge y)]$ the antecedent of the right conjunct (i.e $1 < 0$) is False. Thus, the conditional is True for every value of $y$, irrespective of the value of $x$

Thus the formula $∀y \ (1<0 \to x \ge y)$ is True; but also $0<1$ is, irrespective of the value of $x$.

In conclusion, the original formula is true.

Obviously, its negation will be False, as we can easily verify.