# Poset axioms of Boolean algebra [closed]

I found in Awodey's "Theory Category", second edition p34, a set of poset axioms to define Boolean algebras, whereas I don't see how they can be sufficient.

Here are the axioms:

A Boolean algebra is a poset B equipped with distinguished elements 0, 1, binary operations $a \vee b$ of "join" and $a \wedge b$ of "meet", and a unary operation $\neg b$ of "complementation". These are required to satisfy the conditions $$0 \le a$$ $$a \le 1$$ $$a \le c \text{ and } b > \le c \text{ iff } a\vee b \le c$$ $$c \le a \text{ and } c \le b \text{ > iff } c \le a \wedge b$$ $$a \le \neg > b \text{ iff } a \wedge b = 0$$ $$\neg\neg a =a$$

I can see that from these axioms I can deduce several of the common properties of boolean algebras, so for example I can prove $\forall c, c\wedge\neg c =0$, $\forall (a,c), a\wedge c \le c$, etc. but there are some properties I cannot see how they are provable in this axiomatics.

Here one of which I tried so hard (and failed) to prove from the axioms above: $\forall (a,b), a \le a \vee b$

Could somebody guide me? I am trying to make something impossible? Is my mistake on things I think that are required to be true in a Boolean algebra and are not?

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## closed as off-topic by Ricardo Andrade, David White, Carlo Beenakker, Daniel Moskovich, Ryan BudneyOct 19 '13 at 13:25

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In your second-last axiom, you probably want to say $a\wedge b=0$ rather than $a\vee b=0$. – Joel David Hamkins Mar 19 '13 at 2:44
Yes, thank you. Corrected. I confused three times "vee" and "wedge" while redacting... I hope all is corrected now. – Almeo Maus Mar 19 '13 at 2:48
The claim that $a\leq a\vee b$ seems to follow immediately from the third axiom, by taking $c=a\vee b$, since $c\leq c$. – Joel David Hamkins Mar 19 '13 at 2:52
Exact. Now I feel so stupid that I'd like to close my question... Thank you so much. – Almeo Maus Mar 19 '13 at 2:56
Hey, don't worry about it! We all miss easy things sometimes... – Joel David Hamkins Mar 19 '13 at 3:02

As remarked by Joel, this is indeed very direct starting from $a\vee b \le a \vee b$ and using the third axiom. Many thanks to have unstucked me.