# How to deduce $\square p\to p$ from other modal axioms?

I'm trying to deduce the T axiom $$\square p\to p$$ from the B,D,5 (and also K) axioms.

B: $$q\to\square\diamond q$$

D: $$\square q\to\diamond q$$

5: $$\diamond q\to \square \diamond q$$

I tried to assume $$\square p$$ and $$\neg p$$. By the B axiom, we have $$\square\diamond\neg p$$. By the D and 5 axioms combined, we have $$\square\diamond p$$. Thus we have $$\square\neg(\square p\lor\square \neg p)$$. I've been trying to prove $$\neg\square\neg(\square p\lor\square \neg p)$$ to arrive at a contradiction, but I don't see how to achieve this.

First note that in K if $$\vdash p\rightarrow q$$, then $$\vdash \lozenge p \rightarrow \lozenge q$$. This is because if $$\vdash p\rightarrow q$$, then $$\vdash \lnot q\rightarrow \lnot p$$. By necessitation, $$\vdash \square (\lnot q\rightarrow \lnot p)$$. By K, $$\vdash \square \lnot q \rightarrow \square \lnot p$$. This is equivalent to $$\vdash \lnot \lozenge q \rightarrow \lnot \lozenge p$$, which is equivalent to $$\vdash \lozenge p\rightarrow \lozenge q$$.

The reason for this observation is that from 5, we also get $$\vdash \lozenge\lozenge q \rightarrow \lozenge \square \lozenge q$$.

Now to prove $$\square p\rightarrow p$$, it suffices to prove $$q\rightarrow \lozenge q$$, since substituting $$\lnot p$$ for $$q$$ gives the contrapositive of the statement you want to prove. So assume $$q$$.

By B, $$\square \lozenge q$$. By D, $$\lozenge \lozenge q$$. By the consequence of 5 noted above, $$\lozenge \square \lozenge q$$. And finally by the contrapositive of $$B$$ ($$\lozenge \square p\rightarrow p$$), we have $$\lozenge q$$, as desired.

Here's some motivation for where this argument comes from. Semantically, on Kripke frames, T corresponds to reflexivity, B corresponds to symmetry, D corresponds to the "serial" property (for every world, there is at least one world accessible from it), and 5 corresponds to the "Euclidean" property (if $$w$$ and $$w'$$ are both accessible from $$v$$, then $$w'$$ is accessible from $$w$$). So the claim that T follows from B, D, and 5 corresponds to the claim that every symmetric, serial, Euclidean frame is reflexive.

To argue this, pick a world $$w$$. By seriality, there is a world $$w'$$ accessible from $$w$$. By symmetry, $$w$$ is accessible from $$w'$$. By Euclideanity, since $$w$$ and $$w$$ are accessible from $$w'$$, $$w$$ is accessible from $$w$$. So we've proven reflexivity.

I used this semantic argument as a kind of informal guide to constructing the syntactic argument above. Think of $$q$$ as a property that holds at a world $$w$$. Then to show $$\lozenge q$$, we can show $$w$$ is accessible from itself. We use seriality (D) and symmetry (B) to "move" to another world $$w'$$ from which $$w$$ is accessible ($$\lozenge\lozenge q$$). Then we use Euclideanity at $$w'$$ ($$\lozenge q \rightarrow \square \lozenge q$$, but "inside the diamond", since we're applying the implication at $$w'$$, not at $$w$$). And then we see where we are ($$\lozenge\square\lozenge q$$) and figure out that we have to use symmetry again to "move" back to $$w$$. It's not a perfect analogy or a foolproof method, but this kind of strategy can be useful.