Proof of statement with empty hypothesis

Question

Let's say I have these two propositions:

Prop A. "If $x \in \mathbb{R}$ such that $x^2=-1$ then $3x^2$ is invertible (that means, $3x^2 \neq 0$)"

Prop B. "If $x \in \mathbb{R}$ such that $x^2=-1$ then $3x^2$ is singular (not invertible)"

I want to ask about the correctness of these proofs:

Proof 1.A and 1.B: There is no $x \in \mathbb{R}$ such that $x^2 = -1$, so no matter what the thesis says, the proposition is true, as there will be no counterexample.

Proof 2.A: Suposing $x\in \mathbb{R}$ such that $x^2 = -1$, we have that $3x^2 = 3(-1) = -3 \neq 0$, so proposition $A$ is true.

Proof 3.B: Suposing $x\in \mathbb{R}$ such that $x^2 = -1$, we have that $3x^2 = 3(-1) = -3 \neq 0$, so $3x^2$ is not singular and proposition $B$ is false.

Obviously, they can not all be right. I think Proof 3.B is wrong. ¿But is proof 2.A right? Because it's basically using the same method (ignoring that no such $x$ exists, and using the hypothesis)

Answer 1

I assume that the propositions mean "For all $x$, if $x \in \mathbb{R}$ such that ..." If so, then to prove that Proposition B is false you would have to prove that there exists some $x \in \mathbb{R}$ such that $x^2 = -1$ and $3x^2$ is not singular. Proof 3.B does not do that. So the final phrase "and Proposition B is false" is incorrect.

Answer 2

All of them are right. If $\neg p$ is true then $p\to q$ and $p\to\neg q$ are true.

Example: If $x\in\Bbb R$ such that $x^2=-1$ then I have tons of umbrellas within a bottle of coke that is in the orbit of Mars. Indeed, assume the contrary. Then, by hypothesis, there exists some $x\in\Bbb R$ such that $x^2=-1$. But this is a contradiction, because no such real number exists. So if $x^2=-1$, that thing about the umbrellas is true.

Answer 3

$\neg P$ and $P$ $\implies Q $ for any propositions $P$ and $Q$ this is known as the principle of explosion .