$\lnot\exists x(S(x) \Rightarrow R(x))$ VS $\forall x(S(x) \Rightarrow R(x))$ Without Using De Morgan's Law

Question

I was doing logic exercises the other day and I encountered the following:

Write this statement symbolically and verify your answer using De Morgan's Law:

No squares are rectangles

My solution:

Let $R(x): x$ is a rectangle and $S(x): x$ is a square. Now, we know none of the squares are rectangles; so in other words, there doesn't exist a square that's also a rectangle. I thought the following would be the its equivalent: $\lnot\exists x(S(x) \Rightarrow R(x))$ but I also thought, can't we say all squares are not rectangles and so I wrote the following: $\forall x(S(x) \Rightarrow \lnot R(x))$ but using De Morgan's law, these two are obviously not equivalent and only the latter is right. Could you explain,without using De Morgan's Law, why the first statement is wrong?

Thanks in advance for any help!

Answer 1

No squares are rectangles

is the same as "There is no object that is both a Square and a Rectangle".

Thus, in symbols :

$¬∃x(S(x) \land R(x))$;

This in turn is equivalent to :

$∀x(S(x) \to ¬R(x))$,

and this is exactly : "All squares are not rectangles".

The quantifiers equivalence : $\lnot \exists \equiv \forall \lnot$ is one of the most intuitive logical equivalence : if there is no Black swan, then all swans are not-Black (and vice versa).

The next step needs the propositional equivalence is : $\lnot (p \land q) \equiv (p \to \lnot q)$.

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Let's start with : "All squares are not rectangles".

What about if there is an object $a$ that is a Triangle ? It is neither a Square nor a Rectangle, and thus $S(a) \to ¬R(a)$ is True ($\text F \to \text T = \text T$).

This means that the object $a$ does not contradict the formula $∀x(S(x) \to ¬R(x))$.

What happens with $¬∃x(S(x) \land R(x))$ ? We have that $S(a) \land R(a)$ is False, and thus we have no issue also with $¬∃x(S(x) \land R(x))$.

But what about $¬∃x(S(x) \to R(x))$ ?

We have that $S(a) \to R(a)$ is True, and thus the Triangle $a$ satisfy the formula $∃x(S(x) \to R(x))$.

This means that the existence of the Triangle $a$ contradicts the formula $¬∃x(S(x) \to R(x))$, and this means that this formula is not equivalent to : "All squares are not rectangles", and thus is not equivalent to "No squares are rectangles" either.

Answer 2

$S(x)\Rightarrow R(x)$ is always true when $S(x)$ is false, that is, when $x$ isn't a square. Therefore the statement $\neg \exists x (S(x)\Rightarrow R(x)) $ would imply, among other things, that there is no object which isn't a square. This is clearly not a conseqence of what you want to convey. As long as there exists object that isn't a square, its existance makes the statement $\neg \exists x (S(x)\Rightarrow R(x)) $ false, but it doesn't falsify the statement "no squares are rectangles".