If, as part of proving an implication, you suppose the antecedent - but the antecedent is false in a specific case, can you ignore that case? For example. Say I have a conditional
$$(\frac{x}{x-2} \leq 3 \ \land \ x \geq 2) \implies x \geq 3$$
then clearly, when $x=2$, I have a problem since then my fraction is undefined. 
Now, suppose I wish to prove the above implication strictly by means of a direct proof. That means assuming the antecedent, and from there on trying to prove that it logically leads to the consequent.
However, if I assume the above antecedent $(\frac{x}{x-2} \leq 3 \ \land \ x \geq 2)$, then what do I do about the $\geq$? After all, this means $= \, or \, >$, and since I'm assuming this to be true (as part of the overarching conjunction being true), how do I deal with the conflict of both my antecedent being correct, but $x=2$ being impossible and incorrect?
Is it valid for me to say "Because in the case of x=2, the fraction is undefined, our assumption reduces to $(\frac{x}{x-2} \leq 3 \ \land \ x > 2)$" and then go on to prove $(\frac{x}{x-2} \leq 3 \ \land \ x > 2)$? 
That somehow feels like cheating, or at the least like giving some sort of vacuous/irrelevant proof because I've reduced my assumption (and, by extension, my implication) to something it strictly wasn't. Can somebody shine some light on this?
 A: I think that, since the quantification on $x$ is implicit, you can assume that the $x$ concerned are those for which the assertions $\frac{x}{x-2}\leq 3$ and $x\geq 2$ make sense simultaneously, and what you are trying to prove is more precisely
$$\forall x\in\mathbb{R}\setminus\{2\},\frac{x}{x-2}\leq 3\,\wedge x\geq 2\implies x\geq 3.$$ 
It reminds me a question I asked of the same nature (here), with a nice answer from John M. Lee.
A: You said $(\dfrac{x}{x-2}\leq 3~~ \wedge ~~x\geq 2 )\Rightarrow x\geq 3$. So when you have the antecedent, you get the consequent. But, the thing is that you just have the first inequality of the antecedent if $x \not =2$, if $x=2$ you can't say anything about $\dfrac{x}{x-2}$, because it is not defined as a real number, so you can't compare it with the $\leq$ of $\Bbb{R}$. Since $x \geq2$ iff ($x>2 ~~\vee~~x=2$) there is no problem with the first inequality being satisfied just when $x \not = 2$, because the second inequaliuty can still be satisfied (since $\vee$ requires just one to be satisfied). Therefore, there is no problem with the implication.
