So for starters I want to prove that propositional logic is sound, using a Hilbert system such as Łukasiewicz's.
This system comes with three axioms:
$A \to (B \to A)$
$(A \to (B \to C)) \to ((A \to B) \to (A \to C))$
$(\lnot A \to \lnot B) \to (B \to A)$
And one rule of inference (modus ponens):
$P, (P \to Q) \vdash Q$
To prove soundness I need to show that:
$A \vdash B \implies A \vDash B$
Axiom 1:
$A = \top, B = \top$:
$\top \to (\top \to \top) = \top \to \top = \top$
$A = \top, B = \bot$:
$\top \to (\bot \to \top) = \top \to \top = \top$
$A = \bot, B = \top$:
$\bot \to (\top \to \bot) = \bot \to \bot = \top$
$A = \bot, B = \bot$:
$\bot \to (\bot \to \bot) = \bot \to \top = \top$
And then similar idea for the other two axioms (which I'm not showing here for space purposes since the second axiom has $8$ cases and then the third axiom has $4$ cases again)... but all cases come out $\top$ over all possible inputs of $\top$ and $\bot$ for the propositional variables.
So each axiom is always a true statement, i.e. a tautology (given our definitions/truth tables for $\top$, $\bot$, $\to$, and $\lnot$).
Now I am a little stuck. We still need to show $A \vdash B \implies A \vDash B$, which I believe means we have to somehow show that modus ponens always allows us to derive true deductions, but then somehow using our result that we have true axioms in the form of tautologies.