Is the statement “Every implication or its converse must be true” correct, or is it possible to have both implication and converse be false?
If $A$ stands for "$x$ is a prime number" and $B$ stands for "$x$ is greater than $100$", then both $(A → B)$ and $(B → A)$ are false. Is this a counterexample to the quoted statement? How does this make sense with respect to the truth table of $(A → B) ∨ (B → A)$ showing that it is always true?
Put $Px$ for $x$ is prime, and $Gx$ for $x > 100$.
Now distinguish
from
The first I think properly captures the thought that you say is false (for yes, the generalization $\forall x(Px \to Gx)$ is false, and likewise $\forall x(Gx \to Px)$ is false). But (1) is not a logical theorem.
On the other had (2) is a classical theorem, but is not obviously false, once distinguished from (1). Take any $n$ you like. Then either $Pn$ is false, in which case the supposition that $Pn$ is true leads to contradiction, and anything goes, so we get in particular $(Pn \to Gn)$. Or $Pn$ is true, in which case $(Gn \to Pn)$. So either way we get $((Pn \to Gn) \lor (Gn \to Pn))$. But $n$ was arbitrary so we can generalize.