I am having trouble understanding a certain computation in a proof I am reading.
Suppose $I$ is an ideal in some Boolean algebra. And suppose we have $¬a_0 =a\vee b$ for some $a\leq a_0$ and $b\in I$.
Then $¬a_0 =¬a_0\land ¬a_0 =(a\vee b)\land ¬a_0 =b$.
I am aware of the basic algebraic properties of Boolean algebras, but I just can't see how $(a\vee b)\land ¬a_0 =b$. Could someone please explain this?
I think this works:
$(a \vee b) \wedge \neg a_0 = (\neg a_0 \wedge a) \vee (\neg a_0 \wedge b) = \neg a_0 \wedge b =(a \vee b) \wedge b = b.$
For the second equality, note that $\neg a \wedge \neg a_0 = \neg(a \vee a_0) = \neg a_0$, using $a \leq a_0$. Then,
$$\neg a_0 \wedge a = (\neg a \wedge \neg a_0) \wedge a = 0.$$
This is just spelling everything out very formally, but to eyeball these sorts of calculations it helps to think of everything as set operations ($\wedge = \cap$, $\leq = \subseteq$, etc.) or as logical connectives with $\leq$ as logical implication.