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Let's say I want to prove that:

$A \Rightarrow B$ or $C$.

I would like to know if the following approach to demonstrate the implication above is legitimate (correct):

Assume that $B$ is false and use this and the assumptions ($A$) to arrive at $C$.

What do you think?

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    I think it's fundamentally important that you add parentheses to this expression. – 2017-01-23
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    $A\implies (B \text{or} C)$? – 2017-01-23
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    See here: https://www.physicsforums.com/threads/proving-an-or-statement.366427/ – 2017-01-23
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    You should probably proceed along the lines, "If $B$ is true, then ($B$ or $C$) is true, so the implication $A\implies$($B$ or $C$) is true. If $B$ is false, then ...", arriving at the desired implication in either case. You should explicitly show the implication holds whether $B$ is true or false. – 2017-01-23
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    See the post [conditionals involving disjunctions](http://math.stackexchange.com/questions/2104937/conditionals-involving-disjunctions). – 2017-01-23

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Assuming you mean $$A\Rightarrow (B\lor C)$$

Yes, that is precisely the right approach. If either of your assumptions don't hold, then the statement is true by assumption, since $F\Rightarrow $_____ is always true, independent of the ____, and $T\Rightarrow (T\lor$ ____$)$ is always true, independent of the ____, since it simplifies to $T\Rightarrow T$. Therefore the case you specify is the only case that needs to be proven to prove that the theorem is true.

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    But see my comment on the original question. Some detail is missing. – 2017-01-23
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    @MPW I disagree, see my expanded answer. – 2017-01-23
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    What I mean is that in presenting the proof, OP should at least state something like "If $B$ is true, the implication follows immediately" or perhaps "Without loss of generality, we may assume $B$ is false" to indicate that the alternative has at least been considered. Note that your expanded answer is in fact going through this step. It really can't be omitted if the proof is to be considered complete. – 2017-01-23
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    @MPW oh this is an interesting little confusion. I'm reading the OP as asking if, for a theorem of mathematics of the form $A\Rightarrow (B\lor C)$, is it sufficient to assume $A$ and $\neg B$ and prove $C$. You're reading the OP as asking about proving a logical formula. Whether my expansion is needed for the proof, or an explanation of why the metatechnique works depends on which understanding of the OP one takes. – 2017-01-23
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This kind of doubts can be solved by algebra of logic; I mean in this case: $$ a\implies (b \vee c) $$ iff $$\bar{a}\vee b \vee c$$ iff $$\overline{(a\wedge \bar{b})}\vee c$$ iff $$(a\wedge \bar{b})\implies c$$