double interpretation of Boolean algebra

Question

It bothers me that there seems to be only a flawed double interpretation of Boolean algebra in terms of classes and of propositions. One can say

A∪B indicates union or inclusive disjunction

A∩B indicates intersection or conjunction

A' indicates complement or negation

but when one comes to

A⊂B

one encounters a difficulty: if we say it indicates subclass or implication we have a grammatical problem. In the other pairs above , the class interpretation yields another class ( e.g. A, B and A∪B are all classes) while the propositional interpretation makes them all propositions. However, in the case of A⊂B , both interpretations are sentences. This creates a problem in that , e.g.,

A⊂(B ⊂C)

is grammatical interpreted propositionally but ungrammatical when interpreted in class theory. I tried fixing it by using

A' ∪ B

for

A⊂B

but it doesn’t work (check it with a Venn diagram). Can we fix this?

Answer 1

As "$A\subseteq B$" is a statement about sets and not a set in itself, I prefer to view its logical analogue as "$A\vdash B$", which is a statement about formulas and not a formula itself.

Therefore, in the context of Boolean algebra, "$A\subseteq B$" seem to be more properly translated as "entails" than as "implies".

---

The proper way to view $\to$, or logical implication, is then by $A\to B\equiv \lnot A\lor B$. Hence $\to$ is not a primitive notion in a Boolean algebra, but a defined one. This makes $A'\cup B$ the corresponding set theoretic analogue of $A\to B$.

Answer 2

One immediate difference is that $\subset$ is used to make a claim about sets, while $\cap$ and $\cup$ are operators that work on sets. Indeed, where $A \cap B$ is another set (i.e. $\cap$ is function that takes in two sets and outputs a new set), $A \subset B$ is not a set ... but a claim. So that's why $\subset$ does not have a logical operator counterpart.

Indeed, when you say:

I tried fixing it by using

A' ∪ B

for

A⊂B

that makes absolutely no sense: $A ' \cup B$ is a set, but $A \subset B$ is not a set in the first place.