User talk:Lord Farin/Backup/Definition:Boolean Algebra

(Moved in from Definition talk:Group Action)

Two things:

1. Should the two definitions get their own subpages?
2. I have a genuinely alternative definition (via a poset; also a page Boolean Ring Induces Poset is in the pipeline) for Definition:Boolean Ring. Is the construct available here sufficiently appreciated that I may spread it out? --Lord_Farin (talk) 18:00, 22 August 2012 (UTC)

In order to stay consistent in style, I believe the two definitions should have their own pages, transcluded as ever.

Show what you got for Boolean Ring. Note that we had this conversation before on how to handle definitions. We either settle on Definition A as the definition, and then several iff pages to specify a variety of equivalent statements, or we allow each of these equivalent statements as equally valid definitions and put together a page (or a series of page) proving their equivalence. The latter may be preferable as it is more likely to accommodate expositions from different sources which treat any of the equivalent statements as the definition. So allowing equal status to all definitions is beginning to sound sensible to me. It may be an "added value" that other sites are not in a position to offer. --prime mover (talk) 18:57, 22 August 2012 (UTC)

Apparently, Deskins (I looked the source up via a web search portal and verified) is erratic in his terminology, since I was under the impression to be discussing Boolean ring as defined on WP; this is clearly not the case. This I found out after fruitlessly painstaking efforts to prove certain identities trivial in the poset approach (which I recalled from my BSc. thesis) algebraically. Ultimately, I decided it couldn't be done. See also this WP. My definition is (and this I did prove) equivalent to the def of Boolean algebra as on WP.

Having my own thesis and at least two books I own backing me up (and, but that doesn't really count, WP), I contend that we best proceed by at least adding a disambig on Boolean ring, and separate the Boolean ring and Boolean algebra defs. --Lord_Farin (talk) 21:31, 22 August 2012 (UTC)

Okay that makes sense. I have not encountered a Boolean ring as defined on Wikipedia. So I understand that what we have on PW is what WP defines as a Boolean algebra? If so, then a simple rename and also-known-as should do the trick. And the linking pages will need adjusting, of course. --prime mover (talk) 21:46, 22 August 2012 (UTC)

No, it's more severe: The notions are incompatible. A Boolean ring as we have it isn't even a ring. Neither does it originate from a poset since $*,\circ$ needn't satisfy the absorption laws. --Lord_Farin (talk) 21:48, 22 August 2012 (UTC)

Yes I know what we have is not a "Boolean ring". It even says so on the page itself, that it's not a ring. But it *is* an example of a Boolean algebra, yeah? So what I'm saying is, it can be renamed to Definition:Boolean Algebra (which is currently a redirect to the page Definition:Boolean Ring), and your (and WP's) definition of Boolean Ring can replace what's currently in Definition:Boolean Ring. Or is there something else more subtle than this? --prime mover (talk) 22:15, 22 August 2012 (UTC)

The categories the stuff is ought to be placed in have to be redetermined. But first, let me finish the second definition. --Lord_Farin (talk) 08:27, 23 August 2012 (UTC)

Definition:Boolean Ring has now been replaced with Definition:Huntington Algebra. --prime mover (talk) 06:19, 24 August 2012 (UTC)

I see the redlink Ordering in Boolean Algebra is Lattice Ordering. This can be more accurately phrased by saying that a Boolean algebra is the same thing as a 'complemented distributive lattice' (Goldblatt's words). We'll get there, I will first embark on a journey establishing the links between B and H algebras. --Lord_Farin (talk) 14:00, 24 August 2012 (UTC)

Argh. Please, someone help me about by providing a proof for both distributive laws for $\sqcup$ and $\sqcap$. I have just wasted the whole evening on them. It is required to use $'$ in the derivation since I have a counterexample satisfying axioms $0$ to $3$. --Lord_Farin (talk) 22:21, 24 August 2012 (UTC)

Not tonight, I'm Fridayed out ... --prime mover (talk) 23:12, 24 August 2012 (UTC)

Ok, I think I have it. I will start posting preliminary results. --Lord_Farin (talk) 09:00, 25 August 2012 (UTC)

All I need is a proper name for $(a \sqcup b) \sqcap a' \preceq b$ (though I could also just post it as a corollary to Complement of Join). People with too much time could take on Meet is Associative and Join is Associative. --Lord_Farin (talk) 10:04, 25 August 2012 (UTC)

't Is all done, observe Meet Distributes over Join and Join Distributes over Meet. I am satisfied, off to bed. --Lord_Farin (talk) 22:24, 25 August 2012 (UTC)

Very pretty. I will catch up later - this weekend is mainly dedicated to domestic activities so time is short and attention span limited. --prime mover (talk) 00:15, 26 August 2012 (UTC)

Minor concern: currently HA requires $\bot_S \ne \top_S$ for a BA to be a HA. Suggestion is to remove restraint on HA, so as to allow $e^* = e^\circ$. --Lord_Farin (talk) 22:31, 27 August 2012 (UTC)

Do we have examples of BAs where $\bot_S = \top_S$? Sounds wrong to me, unless $S = \left\{{\bot_S}\right\}$, i.e. the algebra is trivial (is a guess only).

Interestingly, 1964: W.E. Deskins: Abstract Algebra (who presents this as a Boolean ring) insists on $e^* \ne e^\circ$, but 1965: Seth Warner: Modern Algebra (who introduces the term HA) appears not to insist on distinctness. So you could be all right on that. We might need to amend HA but I'd be interested to see why we might need $\bot_S = \top_S$. --prime mover (talk) 05:21, 28 August 2012 (UTC)

The only example is the trivial BA, so it is a minor point. We have $\forall a \in S: \bot_S \preceq a \preceq \top_S$ so if $\bot_S = \top_S$ then $S = \{\bot_S\}$. --06:40, 28 August 2012 (UTC)

If we were to relax the rule that $e^* \ne e^\circ$ in a HA we may be able to demonstrate that the HA is also trivial - in which case we can relax that rule and introduce the concept of a "trivial HA" thence "trivial BA" and so on. I'll leave it to you if you wish. --prime mover (talk) 06:43, 28 August 2012 (UTC)

This apparently is easy:

$a = a \circ e^\circ = a \circ e^* = e^*$

where the last last equality follows from Identity Elements of Huntington Algebra are also Zeroes (which page may need renaming btw). I'm busy right now but will post it later if noone has in the mean time. --Lord_Farin (talk) 06:48, 28 August 2012 (UTC)

With Huntington Algebra Induces Boolean Algebra up, most of the interesting work is done. It remains to verify that the constructions are mutually inverse, which is plainly tedious. Next, we can turn to the different axiomatisations of a HA, i.e. with the Huntington axiom, with the Robbins axiom and of a BA, as 'complemented distributive lattice'. Interestingly, the Robbins axiom was only first proved by a computer, using eight days... --Lord_Farin (talk) 10:00, 29 August 2012 (UTC)

How to denote boolean algebra

Now I look at it, I wonder whether we should use $(S, \sqcup, \sqcap, \preceq)$ to denote a balg after all. It does correlate with the notation used for an ordered structure (without looking it up, or so I believe). But I have a dislike ("doesn't smell right" for including the distinguished elements of such a structure, as they are already accounted for, being part of $S$.

Further thoughts? --prime mover (talk) 10:31, 25 August 2012 (UTC)

It is allowed to denote it as an ordered structure (your suggestion) because it is. However, as this is not imposed (only implied) by the axioms, there should at least be a note somewhere on the page that refers to Boolean Algebra is Ordered Structure. --Lord_Farin (talk) 11:07, 25 August 2012 (UTC)

Yes of course, that goes without saying. --prime mover (talk) 11:29, 25 August 2012 (UTC)