I'd like to find a proof for this proposition: for every a,b ∈ B, a∧¬b=0 if and only if a ≤ b I understand that, in a Boolean algebra, a∧¬a=0 (it is always false), so I suspect I might use that to do the proof, but I can't find how (or I might be completely on the wrong path).
Anybody has an idea how to do such a proof?
Thank you
Note that $a \leq b$ if and only if $a \vee b = b$. We will indeed use that $b \wedge \neg b = 0$, together with a similar fact, namely that $b \vee \neg b = 1$.
Suppose $a \wedge \neg b = 0$, then $$ a \vee b = (a \vee b) \wedge 1 = (a \vee b) \wedge (b \vee \neg b) = \\ ((a \vee b) \wedge b) \vee ((a \vee b) \wedge \neg b) = b \vee ((a \wedge \neg b) \vee (b \wedge \neg b)) = b \vee (0 \vee 0) = b, $$ so $a \leq b$ as required.
For the other direction, if $a \leq b$, then as before $a = a \vee b$. So: $$ a \wedge \neg b = a \wedge \neg(a \vee b) = a \wedge \neg a \wedge \neg b = 0 \wedge \neg b = 0. $$