From an axiom system of propositional logic to an axiom system for a Boolean algebra

Question

A well-known set of axioms for the classical propositional logic is Lukasiewicz's 3rd set:

$A\rightarrow(B\rightarrow A)$

$(A\rightarrow (B\rightarrow C)\rightarrow ((A\rightarrow B)\rightarrow (A\rightarrow C))$

$(\neg A\rightarrow \neg B)\rightarrow (B\rightarrow A)$

How can I turn this system into an axiomatic system for a Boolean algebra? I know that Boolean algebras form the algebraic semantics of the classical propositional calculus, so I guess it is possible, but I would like a detailed answer as per how to do it.

Edit: Following Stefan's comments, suppose that a Boolean algebra $\mathfrak{A}$ satisfies the following axioms, for every $A,B,C$ in $\mathfrak{A}$:

$A\vee B=B\vee A$, $\;\;\;A\wedge B=B\wedge A$

$(A\vee B)\vee C=A\vee(B\vee C)$, $\;\;\;(A\wedge B)\wedge C=A\wedge(B\wedge C)$

$(A\wedge B)\vee B=B$, $\;\;\;(A\vee B)\wedge B=B$

$A\wedge (B\vee C)=(A\wedge B)\vee (A\wedge C)$, $\;\;\;A\vee(B\wedge C)=(A\vee B)\wedge (A\vee C)$

$(A\wedge -A)\vee B=B$, $\;\;\;(A\vee -A)\wedge B=B$

Answer 1

You can't actually accomplish this, but the question still has some merit to it, as I hope I'll make clear in the second paragraph. You can't actually accomplish this, because formal Boolean Algebras, logically speaking, qualify as first-order theories with an equality predicate with all variables quantified. The operations on Boolean Algebras are also not logical functions, but rather operations on sets. The quantification of variables not existing in this propositional calculus is not so much of a problem. We might also ignore the differences between logical functions and operations on sets. We'll also need to join the logical constants for truth and falsity, 1 and 0 respectively as well-formed formulas to the vocabulary of the language. But, even doing all of that, there is no equality predicate in the propositional calculus you've referred to, and thus you can't derive any of the usual axioms for formal Boolean Algebras, since they all involve that equality predicate.

That said though, in Polish notation, it holds that for any well-formed formulas $\alpha$ and $\beta$ in this propositional calculus, if $\vdash$E$\alpha$$\beta$, then tv($\alpha$) = tv($\beta$) where tv stands for a function which maps a well-formed formula to it's truth value after all variables have gotten correctly substituted by 1 or 0. Thus, you might ask how to prove the correlates of the axioms all having the form $\alpha$ = $\beta$, with those correlates denoted as E$\alpha$$\beta$.

To prove any logical equivalence when logical equivalence is not a primitive function, you'll need the symbolic definition of logical equivalence. So, we'll need:

Epq := NCCpqNCqp

The other definitions needed depend on the correlates of the axioms of Boolean Algebra chosen. There exist single axiom systems in the Sheffer stroke for Boolean Algebra, and thus we might only need a single definition:

Dpq := CpNq

Or we might want to prove the Huntington axioms which involve disjunction, conjunction, and negation.

Apq := CNpq
Kpq := NCpNq

In that case the Huntington axioms correspond to:

1. EAxyAyx
2. EKxyKyx
3. EKxAyzAKxyKxz
4. EAxKyzKAxyAxz
5. EAx0x
6. EKx1x
7. EAxNx1
8. EKxNx0

We might prove these by applying the definitions and seeking after the negations. Or since they consist of equivalences, we might prove CCpqCCqpNCCpqNCqp first, and then abbreviate it as CCpqCCqpEpq, and possibly prove other abbreviated formulas first such as CApqCCprCCqrr. We can then prove CAxyAyx, CKxyKyx, CKxAyzAKxyKxz, CAKxyKxzKxAyz, CAxKyzKAxyAxz, CKAxyAxzAxKyz which give us 1. through 4. We can also prove CxAx0, CKx1x, C1AxNx and CKxNx0. But, we also can use CAxNx1, C0KxNx, CAx0x, CxAx0, and CxKx1 to prove 5. through 8. To prove those we might need to interpret 1 as Cpp, and 0 as NCpp, or instead join C0p and Cp1 to our axiom set.

Edit: If the axioms are as you stated, then we need to prove:

1. EAabAba
2. EKabKba
3. EAAabcAaAbc
4. EKKabcKaKbc
5. EAKabbb
6. EKAabbb
7. EKaAbcAKabKac
8. EAaKbcKAabAac
9. EAKaNabb
10. EKANabb