Derivation of implicational propositional axioms Is there a way to subtract and add properties of axioms to generate new axioms?

For example:

{L} = {P S K} // natural deduction

{P S K} = {P H K I} // natural deduction

{S K} = {?} // constructive logic

{K I} = {?}

Where:

L = ((A -> B) -> C) -> ((C -> A) -> (D -> A)) // Łukasiewicz's axiom system

P = ((A -> B) -> A) -> A // Piece's Law

H = (A -> B) -> ((B -> C) -> (A -> C)) // Weak hypothetical syllogism

S = (A -> (B -> C)) -> ((A -> B) -> (A -> C))

K = A -> (B -> A) 

I = A -> A

I want to be able to be able to add and subtract axioms such that 
P + S + K = P + H + K + I implies S encodes the properties of H + I

I'm probably using unjustified assumptions here. For example, I assume you can derive S from H and I, without using P or K. Ideally, there would be a way to automate the process of constructing and destructing axioms (though it'd probably be NP hard).
The natural way to define your addition operation is as set intersection. The value of a single axiom is the set of all models satisfying the axiom. The sum of several axioms is then the set of all models satisfying all the axioms.

A similar definition is $\mathbf{A} + \mathbf{B} = \mathbf{A} \land \mathbf{B}$, but then you need to interpret equality as logical equivalence. The definitions then become completely equivalent.

However, you can't "cancel" axioms. If you know that $\mathbf{A}+\mathbf{B} = \mathbf{A}+\mathbf{C}$ then all you can conclude is that in the presence of $\mathbf{A}$, $\mathbf{B}$ and $\mathbf{C}$ are equivalent. As an example, let $\mathbf{A}$ be $x=0$, let $\mathbf{B}$ be $y=1$, and let $\mathbf{C}$ be $x+y=1$. In the presence of $\mathbf{A}$, the axioms $\mathbf{B}$ and $\mathbf{C}$ are equivalent, but in general they aren't.

If are are really keen on the cancellation law, you can always take formal sums. It is then true that $\mathbf{A}+\mathbf{B} = \mathbf{A}+\mathbf{C}$ implies $\mathbf{B}=\mathbf{C}$, but at the cost of $\mathbf{A}+\mathbf{B}$ having no other meaning than the formal sum of $\mathbf{A}$ and $\mathbf{B}$.