$x^2=4 \implies \exists x{=}2(x^2=4)$

Question

$x^2=4 \implies x=2\tag1$ $x^2=4 \implies \exists x{=}2(x^2=4)\tag2$

The first proposition is false, since it is logically equivalent to $x\neq2 \implies x^2\neq4$ which is obviously false.

It seems that the second proposition is also true. How can we prove it?

Answer 1

The proposition $\exists x=2(x^2=4)$ is short for $$ \exists x(x=2 \wedge x^2=4)$$

Thus, the second proposition is true. (The '$\wedge$' symbol means "and".)

Answer 2

Try to interpret it in English. in the sense the second statement simply states "If $x^2=4$ then it implies there exists a $x = 2$ such that $x^2=4$".
Which is obviously true, as the statement speaks about being "a $x$" and not a "unique $x$"

Answer 3

Given any propositions (or propositional functions $P(x)$ and $Q(x)$, the conditional $P(x) \rightarrow Q(x)$ is false when $P(x)$ is true and $Q(x)$ is false, otherwise $P(x) \rightarrow Q(x)$ is true.

Now let us put $P(x)$ to be the propositional function $x^2=4$ and $Q(x)$ be the propositional function $\exists x \left( x=2 \land x^2 = 4\right)$. Then as $Q(x)$ is true, so is $P(x) \rightarrow Q(x)$.