Question about implication I have question about something i don't understand.
$\alpha$, $\beta$, and $\gamma$ are statements.
if $\alpha\implies\beta\lor\gamma$ then it's necessary that $\alpha\implies\beta$ or $\alpha\implies\gamma$. that answer is "no" but i can't understand why.
if it's given that $\alpha\implies\beta\lor\gamma$ then it means that when $\alpha$ is true then $\beta\lor\gamma$ must be also true which means that either $\beta$ is true or $\gamma$ is true. so it means that $\alpha\implies\beta$ or $\alpha\implies\gamma$, isn't it? because the only case when it's might be not true is when $\alpha$ is true, $\beta$ is false and also $\gamma$ is false.
So i can't understand why the answer to this question is "no".
I hope someone can explain it.
Thank you. (P.S. sorry for my bad English).
 A: This answer is closely related to Peter Smith's, but from a somewhat different point of view, trying to show what's common to the examples in the various answers.
Notice first that, as Peter pointed out in the second part of his answer, $\alpha\implies(\beta\lor\gamma)$ is tautologically equivalent to $(\alpha\implies\beta)\lor(\alpha\implies\gamma)$. Things begin to go wrong when these formulas are put into a context that explicitly or implicitly involves a universal quantifier and one tries to distribute this quantifier across the $\lor$ connective.  
Thus, in Alex Kruckmnan's example, we do have
$$
\forall n\,[(n\text{ is a natural number}\implies n\text{ is even})\lor
(n\text{ is a natural number}\implies n\text{ is odd})],
$$
but we cannot infer 
$$
[\forall n\,(n\text{ is a natural number}\implies n\text{ is even})]\lor
[\forall n\,(n\text{ is a natural number}\implies n\text{ is even})].
$$
Similarly, in Nathan Smith's example, the trouble arises from the universal quantifier implicit in "necessarily", which means "in all possible situations".  Under the stated hypotheses, it is true that "necessarily, [if there is traffic then it is the morning rush hour or if there is traffic then it is the evening rush hour]", but it is not true that "[necessarily, if there is traffic then it is the morning rush hour] or [necessarily, if there is traffic then it is the evening rush hour]".
Similarly, in the first paragraph of Peter Smith's answer, the notion of logical validity includes an implicit universal quantifier "for all truth assignments" and the problem arises from attempting to distribute this quantifier over a disjunction.
A: Consider the following statements:
$\alpha$: $n$ is a natural number.
$\beta$: $n$ is even.
$\gamma$: $n$ is odd.
Certainly $\alpha\implies \beta\lor\gamma$, since if $n$ is a natural number, $n$ is even or $n$ is odd.
But we don't have $\alpha \implies \beta$ or $\alpha \implies \gamma$, since not every natural number is even and not every natural number is odd.
A: My summary of the two interpretations of your question:
Is $(a \Rightarrow (b \vee c)) \Rightarrow ((a \Rightarrow b) \vee (a \Rightarrow c))$ a tautology?
Yes.
If $a \vdash b \vee c$ then does $ a \vdash b$ or  $a \vdash c$?
No.
A: Ok.  Just use an example:
If there is traffic it implies it is the morning rush hour or the evening rush hour.
What this statement says it that if there is traffic, it could be either the morning or the evening.... but it's not necessarily either.
So the statements
traffic implies it's the morning or traffic implies it's the evening aren't true.
A: *

*Put $\alpha = (P \lor Q)$, $\beta = P$, $\gamma = Q$. Then $\alpha \vDash \beta \lor \gamma$, trivially, but neither $\alpha \vDash \beta$ nor $\alpha \vDash \gamma$ (where '$\vDash$' as usual indicates logical entailment, and $P$, $Q$ are contingent). So the principle fails when $\Rightarrow$ is read as $\vDash$, i.e. as logical entailment. (Exactly similarly if it is read as $\vdash$, deducibility in a given proof system.)

*Suppose neither $\alpha \supset \beta$ nor $\alpha \supset \gamma$ (where '$\supset$' is the material conditional). Then we'd have to have $\alpha$ true and both $\beta$ and $\gamma$ false, and hence  $\alpha \supset \beta \lor \gamma$ would be false too. Contraposing, necessarily, if $\alpha \supset \beta \lor \gamma$ is true, so is one of $\alpha \supset \beta$ and $\alpha \supset \gamma$. So the principle indeed holds when $\Rightarrow$ is read as the material conditional (and the necessity takes wide scope). 
So, crucially, you need to be clear about the reading you are supposed to be giving $\Rightarrow$ which (most unfortunately) is used in different ways in different texts.
