So here goes.

A little primer on modal logic.

A quick-and-dirty explanation of logical operators:

(p AND q) means p is true and q is true

(p OR q) means p is true, q is true, or both p and q are true.

~p is a proposition that has the opposite truth value of p.

USEFUL PROPERTIES

~~p = p

DeMorgan's laws:

~(p AND q) = (~p OR ~q)

I.e., if it's false that "John is tall AND Susie is tall", either John or Susie or both must not be tall.

~(p OR q) = (~p AND ~q)

I.e., if it's false that "He ate cake OR he ate ice cream" then he couldn't have eaten cake AND he couldn't have eaten brownies.

So you kind of "distribute" the ~ through by negating each proposition and flipping OR with AND (and vice versa). It's kind of ugly with nested operators (i.e., A AND (B OR (C AND D))), but it can be done by continuing to apply the rule.

You can also do it in reverse:

(p AND q) = ~~(p AND q) = ~(~p OR q)

[This is similar to saying (a + b) = (-1)*(-1)*(a + b) = (-1)*(-a + -b)]

I'll use N(q) for the proposition "Necessarily, q is true", which is equivalent to "q is true in every possible world". I'll use P(q) for the proposition "Possibly, q is true", which is equivalent to "There exists at least one possible world in which q is true."

PROPERTIES OF N(x) AND P(x)

N(x) = ~P(~x)

Proof: If x is true in every possible world, then there can't be a world where it is false. Thus there does not exist a possible world where ~x holds. This is the definition of ~P(~x).

Conversely, suppose that there are no possible worlds in which ~x holds. Then there are no possible worlds in which x is false. Thus x is true in every possible world. This is simply N(x)

P(x) = ~N(~x)

Suppose there is a possible world in which x is false; call it W. If N(~x) were true, ~x would hold in every possible world. In particular, ~x would hold in W. But x holds in W; this is a contradiction. Hence N(~x) must be false, thus ~N(~x) holds.

Conversely, Suppose ~N(~x) holds. Then it is false that ~x holds in every possible world.

DeMorgan's Modal Laws:

~(N(p) AND N(q)) = ~N(p) OR ~N(q)

Suppose "N(p) AND N(q)" is false. That is, "p is true in every possible world AND q is true in every possible world" is false.

Case 1: p is true in every possible world, but q isn't. Then N(q) is false. Hence ~N(q) is true. Thus ~N(p) OR ~N(q) is true.

Case 2: p is true in every possible world, and so is q. Thus N(p) AND N(q). This contradicts our first assumption.

Case 3: p isn't true in every possible world, but q is. Thus ~N(p). Hence ~N(p) OR ~N(q) is true.

Case 4: p isn't true in every possible world, and neither is q. Thus ~N(p) (and also ~N(q)). Hence ~N(p) OR ~N(q) is true.

In any case not entailing contradiction, ~N(p) OR ~N(q) is true.

Conversely, suppose ~N(p) OR ~N(q).

Case 1: ~N(p). Thus N(p) is false. Suppose "N(p) AND N(q)" is true; then N(p) is true. Contradiction. Thus ~(N(p) AND N(q)) holds.

Case 2: ~N(q). Thus N(q) is false. By similar argument above, ~(N(p) AND N(q)) holds.

In either case, ~(N(p) AND N(q)) holds.

QED.

LEMMA 1

~(P(x) AND P(y)) = ~P(x) OR ~P(y)

Proof:

~(P(x) AND P(y))

= ~(~N(~x) AND ~N(~y)) [Since P(x) = ~N(~x)]

= N(~x) AND N(~y) [By DeMorgan's modal law]

= ~P(x) AND ~P(y)

-> ~P(x) OR ~P(y)

Conversely,

~P(x) OR ~P(y)

= N(~x) OR N(~y)

Case 1: N(~x). Then ~x holds in every possible world. Now suppose "P(x) AND P(y)" is true. Then there is a possible world in which x is true; this contradicts N(~x). Hence "P(x) AND P(y)" must be false, i.e. ~(P(x) AND P(y)) holds.

Case 2: N(~y). By similar argument above, ~(P(x) AND P(y)) holds.

In either case, ~(P(x) AND P(y)) holds.

QED

LEMMA 2

N(p AND q) -> N(p) AND N(q)

Proof: Suppose N(p AND q). Then "p AND q" is true in every possible world. In particular, p is true in every possible world. Thus N(p) holds. By similar argument, N(q) holds. By definition of AND, N(p) AND N(q) holds.

LEMMA 3

P(p AND q) => P(p) AND P(q)

Proof: Suppose P(p AND q). Then there exists at least one possible world where "p AND q" holds. Thus there exists at least one possible world where p holds, hence P(p) is true. By similar argument, P(q) holds. By definition of AND, P(p) AND P(q) holds.

THE MAIN QUESTION

Is the following true?

N(p OR q) => N(p) OR N(q)

It would seem not; just because "p OR q" must be true in every possible world shouldn't mean that p is true in every possible world, or q is true in every possible world. For example, take p = "There exists at least one human", and q = "There does not exist any humans". Then in any possible world, one of {p,q} will be true; thus, N(p OR q).

But clearly N(q) is false, since the actual world has humans and is possible. And N(p) isn't obviously true, since we can imagine worlds where humanity never evolved.

THE RUB

N(p OR q)

= N(~(~p AND ~q) [By reverse DeMorgan's law]

= ~P~(~(~p AND ~q) [Since N(x) = ~P(~x)]

= ~P(~p AND ~q) [since ~~x = x]

-> ~(P(~p) AND P(~q)) [By lemma 3]

-> ~P(~p) OR ~P(~q) [By lemma 1]

= N(p) OR N(q) [since ~P(~x) = N(x)]

QED

This has strange implications. For instance, suppose p is a proposition true in the actual world. Consider "p OR ~p". This must be true in every possible world; hence N(p OR ~p). Then by the above argument, N(p) OR N(~p). If N(~p) were true, then p would be false in every possible world, including ours--a contradiction. Hence N(p) must be true.

That is,anything true in our world is true in.anypossible world

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