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