Question about proofs using the converse I'm reading the MCS textbook from MIT
where they state the following:

The valid formula of part (a) leads to sound proof method: to prove
that an implication is true, just prove that its converse is false.
For example, from elementary calculus we know that the assertion "If a
function is continuous, then it is differentiable" is false. This
allows us to reach at the correct conclusion that its converse, "If a
function is differentiable, then it is continuous" is true, as indeed
it is.


But wait a minute! The implication "If a function is differentiable,
then it is not continuous" is completely false. So we could conclude
that its converse "If a function is not continuous, then it is
differentiable", should be true, but in fact the converse is also
completely false.


So something has gone wrong here. Explain what.

Note: the formula of part (a) is "$(P \text{ implies } Q) OR (Q \text{ implies } P)$ is true"
Now I am having trouble solving this question. On the Wikipedia page for Converse it states that "the truth of the converse is generally independent from that of the original statement", so I initially thought the answer was, "the converse is not necessarily related to the original statement" and left it that. But then that is at odds with the formula of part (a) as well as the claim the textbook makes that you can prove something using the converse.
Any help would be very appreciated!
 A: This is a pretty confusing question to ask early in one’s logical education! What’s going on is that, if we want to think about the statement “$p$ implies $q$” and we can prove that the converse “$q$ implies $p$” is false, then in fact we know that $q$ is true and $p$ is false, which means $p$ does imply $q$. That is, symbolically, $$\neg(q\to p)\to (p\to q).$$
But this last implication is not an equivalence! That is, knowing $p\to q$ is true doesn’t mean $q\to p$ is false, or equivalently, knowing $q\to p$ is true doesn’t mean $p\to q$ is false.
Thus a statement is to some extent independent of its converse, but not perfectly so, as it’s impossible for both a statement and its converse to be false. This comes somehow from the asymmetry of the truth table for implication, which has three T’s out of four.
A: If you translate $[A \implies B]~~$ to $~~[(\neg A) \vee B]~~$ then the assertion that (in general) $~~(P \implies Q)~~$ must hold when $~~\neg(Q \implies P)~~$ ::: is a true assertion, demonstrated as follows:
The only way that the statement $(Q \implies P)$ can be false is if
$$(\neg P) \wedge Q.\tag1$$
The only way that the statement $(P \implies Q)$ can be false is if
$$(\neg Q) \wedge P.\tag2$$
If statement (1) above is true, then statement (2) above can not be true.
