Right Anti-Distributivity of Implication over Conjunction There are four unpopular rules of replacement concerning the relationship between Implication (conditional, "if-then" statement, $\implies$) and Disjunction ("or", $\lor$) and Conjunction ("and", $\land$). Below, the symbol $\dashv\vdash$ stands for logical equivalence.
The first two laws describe left-distributivity, and are sometimes called Laws of Composition:


*

*Implication is Left Distributive over Disjunction
$$p \implies (q \lor r) \dashv\vdash (p\implies q) \lor (p \implies r)$$

“If I’m working, then I’m listening to music or I’m sitting down.” is equivalent to “If I’m working, then I’m listening to music, or, if I’m working, then I’m sitting down.”


*Implication is Left Distributive over Conjunction
$$p \implies (q \land r) \dashv\vdash (p\implies q) \land (p \implies r)$$

“If I’m working, then I’m thinking and I’m not eating.” is equivalent to “If I’m working, then I’m thinking, and, If I’m working, then I’m not eating.”

The right-hand laws are anti-distributive, in the sense that the dual operators, "and" and "or", are switched for one another in the conclusion.


*Implication is Right Anti-Distributive over Disjunction
$$(p \lor q)\implies r \dashv\vdash (p\implies r) \land (q \implies r)$$

“If I’m in NY or I’m in LA, then I’m in the USA.” is equivalent to “If I’m in NY, then I’m in the USA, and, if I’m in LA, then I’m in the USA.”


*Implication is Right Anti-Distributive over Conjunction
$$(p \land q)\implies r \dashv\vdash (p\implies r) \lor (q \implies r)$$

 “If I’m in NY and I’m in LA, then I’m dead.” is equivalent to “If I’m in NY, then I’m dead, or, if I’m in LA, then I’m dead.”

The first three make sense to me, but I need some help with this last one. I understand #4 technically, and I’ve verified that it’s true using truth table, Venn diagram, and Natural Deduction. It’s hard for me to understand intuitively though, especially since the sentences in quote #4 do not seem equivalent: if I’m in both NY and LA at the same time, then I’m obviously dead (ignoring paranormal and quantum physical possibilities), however I can very well be in NY while remaining perfectly intact, or, in LA while remaining perfectly intact. Where’s the flaw in my thinking?

Verification of #4 via truth table:
p | q | r | p&&q | p->r | q->r | (p&&q) -> r | (p->r) || (q->r)
--|---|---|------|------|------|-------------|-----------------
0 | 0 | 0 |  0   |  1   |  1   |        1    |        1
0 | 0 | 1 |  0   |  1   |  1   |        1    |        1
0 | 1 | 0 |  0   |  1   |  0   |        1    |        1
0 | 1 | 1 |  0   |  1   |  1   |        1    |        1
1 | 0 | 0 |  0   |  0   |  1   |        1    |        1
1 | 0 | 1 |  0   |  1   |  1   |        1    |        1
1 | 1 | 0 |  1   |  0   |  0   |        0    |        0
1 | 1 | 1 |  1   |  1   |  1   |        1    |        1

via Natural Deduction:
(p && q) -> r          --- premise
!(p && q) || r         --- dfn Implication
(!p || !q) || r        --- De Morgan
(!p || r) || (!q || r) --- Disjunction Distributivity over Disjunction (see below＊)
(p -> r) || (q -> r)   --- dfn Implication. conclusion

(*mini-proof of OR-distributivity:)
(!p || !q) || r        --- premise
(!p || !q) || (r || r) --- Disjunction Idempotency
!p || (!q || (r || r)) --- Disjunction right-Associativity
!p || ((!q || r) || r) --- Disjunction left-Associativity
!p || ((r || !q) || r) --- Disjunction Commutativity
!p || (r || (!q || r)) --- Disjunction right-Associativity
(!p || r) || (!q || r) --- Disjunction left-Associativity. conclusion

 A: That equivalence is a perfect example of the Paradox of Material Implication, which stems from the fact that the material conditional (i.e. the mathematically defined truth-function that works the way it does as given by its truth-table) does not perfectly match up with the 'if ... then ..'  conditionals we use in natural language and thus in everyday thinking.
Before getting to your case, here is another example of this mismatch: consider the statement 'If John lives in Los Angeles, then John lives in New York'. Now, any normal person would consider that statement to be false, regardless of where John actually lives. But note: if John does not live in Los Angeles, then the 'if' part is false, so if we were to use the truth-table of the material conditional, the statement would be true!
OK, so did we choose the wrong truth-values in our truth-table definition for the conditional, e.g. should maybe the conditional be set to false once the antecedent is false? No, because if you say 'If John lives in Los Angeles, then John lives in California' should clearly be true.
So, as it turns out, we can give all kinds of examples where our intuitions regarding the truth of an every day conditional either agrees or disagrees with the truth-table definition, i.e. the material conditional. For several good reasons, the truth-table definition that we did pick is, out of possible truth-table definitions, the best one ... But it isn't perfect: our English conditional is often used in a way that is not truth-functional at all! Hence the 'Paradox of the Material Implication'
Your particular example 'suffers' from this Paradox as well. Consider:
$P$: ' John is a (adult) man'
$Q$: 'John is unmarried'
$R$: ' John is a bachelor'
With these, the left hand side makes perfect sense ... but the right hand side not!
The moral: you have to be careful with using the material conditional to analyze real life conditionals. Oftentimes, the analysis is unproblematic and in accordance with our intuitions, but sometimes it is not!
A: The answer by Bram28 already discussed that the problem is a consequence of how material implication is usually defined.
You might find a different take on implication more satisfying, which becomes possible once we shift away from classical logic to a different logic. Implication can then have a different meaning, which can eliminate the problem.
Your example shows that the problematic part of the 4. law is
$$
 (A \land B) \to C \vdash (A \to C) \lor (B \to C)
$$
and indeed e.g. in intuitionistic logic this statement is not provable$^1$, which immediately resolves the paradox.
In intuitionistic logic we get a different equivalent statement to $(A \land B) \to C$, namely
$$
  (A \land B) \to C \dashv \vdash A \to (B \to C)
$$
which of course also holds in classical logic.
(One might also recognize this a principle from programming called currying)

$(^1)$ By choosing $A := A$, $B := \neg A$ and $C := \bot$ it could be used to prove weak excluded middle.
