Modal logic - Show that if $\vdash \Diamond T$ holds, $\vdash \Box A \to \Diamond A$ holds

Question

In normal Modal logic, how can I show that if $\vdash \Diamond T$ holds (is derivable), $\vdash \Box A \to \Diamond A$ also holds. I can already prove it by showing that if $\vdash \Diamond T$ holds the frame must be serial and then showing that if a frame is serial $\vdash \Box A \to \Diamond A$ also holds. But how can I show it by deduction? Any hints?

Answer 1

We can use the following propositional tautology $$\vdash (A \land \neg A)\rightarrow \neg \top$$

Then using inverse deduction rule, necessitation rule and deduction we obtain $$\vdash \Box(A \land \neg A)\rightarrow \Box \neg \top$$

Then by using the fact that $\Box$ distributes over $\land$ we get:

$$\vdash (\Box A \land \Box \neg A) \rightarrow \Box \neg \top$$

By using contraposition we get:

$$\vdash \neg \Box \neg \top \rightarrow \neg(\Box A \land \Box \neg A)$$

Now using DeMorgan to get $$\vdash \neg \Box \neg \top \rightarrow \neg \Box A \lor \neg \Box \neg A)$$

Or, equivalently:

$$\vdash \Diamond \top \rightarrow \neg \Box A \lor \Diamond A$$

Now rewriting the $\lor$ in terms of implication we finally obtain:

$$\vdash \Diamond \top \rightarrow ( \Box A \rightarrow \Diamond A )$$

Now we can use the fact that we have $\vdash \Diamond \top$ and the cut rule in order to get $\vdash ( \Box A \rightarrow \Diamond A )$

Answer 2

Hilbert style: from $\vdash \Diamond T$ infer $\vdash \Diamond (\neg A\vee A)$ by Boolean equivalence, then $\vdash \Diamond \neg A \vee \Diamond A$ by $\Diamond$-addition, so by operator duality $\vdash \neg\Box A \vee \Diamond A$ i.e. $\vdash \Box A \rightarrow \Diamond A$. In your system you should be able to derive what you need from this trick.