Why when the implication is false, its converse is also false in this example?

Question

As we all know:

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. But where?

Answer 1

It is logically correct to say that $$ (p \implies q)\quad OR \quad (q \implies p)$$ It is also logically correct to say $$ [(p(x)\implies q(x))\quad OR \quad (q(x) \implies p(x))] \quad \forall x \in X$$ However, it is not logically correct to say that $$ [(p(x)\implies q(x)) \quad \forall x \in X] \quad OR \quad [(q(x)\implies p(x)) \quad \forall x \in X] $$

Your statement about differentiable functions has the latter form (and is not correct):

$$[(\mbox{$f$ differentiable} \implies \mbox{$f$ not continuous})\forall f \in F]\quad OR \quad [(\mbox{$f$ not continuous} \implies \mbox{$f$ differentiable})\forall f \in F]$$ where $F$ is the set of all functions from $\mathbb{R}$ to $\mathbb{R}$. On the other hand, if we take any specific function $f \in F$, such as $f(t) =e^t$ for all $t\in \mathbb{R}$, then we indeed get a correct statement: $$(\mbox{$e^t$ differentiable} \implies \mbox{$e^t$ not continuous})\quad OR \quad (\mbox{$e^t$ not continuous} \implies \mbox{$e^t$ differentiable})$$ and indeed the latter is true.

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Aside: Proof of the more general logical statement: For any three propositions $p,q,r$ we have $$ (p\implies q) \quad OR \quad (q \implies r)$$

Proof: \begin{align} (p\implies q) \quad OR (q \implies r) &= (\lnot p \quad OR \quad q) \quad OR \quad (\lnot q \quad OR \quad r)\\ &= (\lnot p) \quad OR\quad (q) \quad OR \quad (\lnot q) \quad OR\quad r\\ &= True \end{align}

Answer 2

Your claim isn't exactly correct, it usually known this way, if you want to prove $p\implies q$ (i.e. if $p$ then $q$) you can prove that $\lnot q \implies \lnot p$ (i.e. if $q$ is false then $p$ is false), they are both the same thing.