Modal Logic: Equivalence between two formulas

Question

I want to know if these two modal formulas are equivalent: $\forall x (B(x) \to \diamond (B(x) \land \lnot x))$ $\equiv$ $\forall x (B(x) \to \diamond \lnot x)$.

The -> direction holds and I have a proof. But what about the <- direction? Can someone give a proof or tell why it does not hold?

Just for the context: Both formulas want to express philosophical skepticism. The first formula reads something like: for all propostions x it holds that if you believe in x then it is possible that you believe in x but x is false, the second formula says: for all propositions x it holds that if you believe in x then it is possible that x is false.

Answer 1

There are two versions of this. In both versions, the two are not logically equivalent.

Version 1: we're in modal first-order logic. There are two predicates: $B(x)$ and $T(x)$, where $T(x)$ informally means "$x$ is true" and $B(x)$ informally means "I believe $x$". The statements are $\forall x , B(x) \to \diamond (B(x) \land \neg T(x))$ and $\forall x , B(x) \to \diamond \neg T(x)$.

In this version, there is no equivalence between the two. To prove this, consider a scenario where there is only one $x$ being quantified over and there are two possible worlds. In possible world $A$, both $B(x)$ and $T(x)$ hold, while in possible world $B$, both $B(x)$ and $T(x)$ do not hold.

So in possible world $A$, we see that $\forall x, B(x) \to \diamond(B(x) \land \neg T(x))$ fails. By contrast, $\forall x, B(x) \to \diamond \neg T(x)$ holds in all possible worlds.

So the two are not equivalent.

Version 2: we're some version of higher-order modal logic, and $x$ is actually a proposition variable.

Define $B(x) = x \land \diamond \neg x$. We see that $B(x) \to \diamond \neg x$ is a tautology, and thus $\forall x, B(x) \to \diamond \neg x$ is a tautology.

We also see that $B(x) \land \neg x$ is always false by definition. That is, $\neg \diamond (B(x) \land \neg x)$.

However, suppose we have some proposition $P$ which holds in possible world $A$ but does not hold in possible world $B$.

Then $\diamond \neg P$ holds. So in possible world $A$, we have $P \land \diamond \neg P$; that is, we have $B(P)$. But we've already show that $\neg \diamond (B(P) \land \neg P)$.

Thus, the statement $\forall x, B(x) \to \diamond (B(x) \land \neg x))$ does not hold in possible world $A$.

So again, the two statements are not equivalent.

Answer 2

Here is my attempt to prove the equivalence $\forall x (B(x) \to \diamond (B(x) \land \lnot x))$ $\equiv$ $\forall x (B(x) \to \diamond \lnot x)$.

Again, I just prove <- direction since the other way around got proved already.

So let's assume $\forall x (B(x) \to \diamond \lnot x)$ to be true. Then we have 3 possible cases:

1. $B(x)$,$\diamond \lnot x$
2. $\lnot B(x)$, $\diamond \lnot x$
3. $\lnot B(x)$, $\square x$

In cases 2 and 3 both upper formulas are trivially true, so case 1 is crucial. In case 1 we'd have a (possible) world w with $\lnot x$. Since B(x) ought to be true as well we can always assume another possible world w‘: w $\cup$ B(x), so it's true that $\diamond (B(x) \land \lnot x)$. But now also $\forall x (B(x) \to \diamond (B(x) \land \lnot x))$ becomes true by verum ex quodlibet.

So whenever $\forall x (B(x) \to \diamond \lnot x)$ is true, so is $\forall x (B(x) \to \diamond (B(x) \land \lnot x))$ which means nothing less than $(\forall x (B(x) \to \diamond \lnot x)) \to (\forall x (B(x) \to \diamond (B(x) \land \lnot x)))$.

Opinions?

Answer 3

It sounds like you're using a proprietary variant of doxastic logic where you can quantify over propositions and further mixing with $\diamond, \square$. But your proof is invalid under the most general normal system K regarding your above "In case 1 we'd have a (possible) world with $\lnot x$ and $B(x)$". You can only have a possible world with $\lnot x$, but cannot derive you also have $B(x)$ in the same possible world. Just imagine if your current world $w$ is non-reflexive, and at all accessible worlds you don't believe $x$ (except at current world $w$), then obviously you cannot derive the left-hand side conjunction starting with $\diamond$...