Boolean logic to arithmetic is it possible to convert a Boolean logic expression to an arithmetic equation?
For example, I have the following equation: T = C $\cup$(B $\cap$ A)
The above equation obeys the distributive law, so can I convert it so:
C $\cup$(B $\cap$ A) = C $\times$B + C $\times$ A?
No, rather, the distribution of union over intersection gives us:
$$C\cup (B\cap A) = (C\cup B) \cap (C\cup A)\tag{Distribution}.$$
So we can choose to prove this in logic, by showing  $$x\in [C\cup (B\cap A)] \equiv x \in [(C\cup B)\cap (C\cup A)]$$
so $$ [x\in C \lor x \in (B\cap A)] \equiv [x\in (C\cup B)] \land [x \in (C\cup A)]$$
...
No, rather, the distribution of union over intersection gives us:
$$C\cup (B\cap A) = (C\cup B) \cap (C\cup A)\tag{Distribution}.$$
So we can choose to prove this in logic, by showing  $$x\in [C\cup (B\cap A)] \equiv x \in [(C\cup B)\cap (C\cup A)]$$
so $$ [x\in C \lor x \in (B\cap A)] \equiv [x\in (C\cup B)] \land [x \in (C\cup A)]$$
...