Martin Howe Posted May 14, 2019 (edited) Alvin Plantinga argues ◊□X → □X is not counter-intuitive and gives a possible-worlds justification for it. But all that does is *prove* it; that doesn't show how it is *intuitive* in the way that X → □◊X is. Can anyone more knowledgeable than me give me a *practical* example, with a contradiction that results if it were false? For the curious: Spoiler This is about modal logic. I was hoping some of my fellow doomers would know the answer as it's stumped even Google. The diamond symbol means "possibly true", in other words it could or could have happened even if it never actually does/did. The square symbol means "necessarily true", in other words it must be always true and can't ever not be. The arrow means "implies". So X → □◊X means "If something actually happened, it *must* be *possible*" ("necessarily" is represented here by the word "must"). This is easy to understand without maths as it is obvious with a bit of thought ("intuitive"). It's known as Brouwer's modal axiom, abbreviated to (B). But ◊□X → □X is problematic. It means "if it is *possible* for something to be necessarily true, then it *is* necessarily true". This is a modal logic equivalent of time dilation; it is mathematically sound, but makes no sense intuitively. The proof by possible-worlds theory is valid, but that does not make the equation seem like common sense, only shows its true. Some logical systems avoid defining this and use Brouwer's axiom instead, treating the other one as "Oh well, it exists but we'll avoid it using it where possible". It's a bit like those classes in .NET marked "This is infrastructure and should not be called directly from your code". (In other words, Beware of the Leopard :p ) Indeed, it's not definitely obvious that there are even valid meanings for things like "necessarily possible" and some mathematicians reject composing the "necessary" and "possible" operators at all. 0 Share this post Link to post
andrewj Posted May 15, 2019 An implication is only false if the LHS is true and the RHS is false. The RHS being false could be phrased as "X is not necessarily true". Since X *is* not necessarily true, then there is no possibility of X being necessarily true, hence the LHS must be false as well. So the implication holds. Assuming the above is right (I really don't know), what makes it un-intuitive for me is the implication operator. 0 Share this post Link to post
Martin Howe Posted May 15, 2019 Thanks. In fact, I seem to remember that one of the reasons for the existence of modal logic in the first place was the fact that the classical definition of implication took no account of any intentional context. 0 Share this post Link to post