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@Laurent Meesseman: I reverted your edit but I had another look and the result is confusing. My reasoning was based on the following:

x+y  means  x XOR y
xy   means  x AND y

  x  y   x+y  xy   x+y+xy
  -----------------------
  0  0    0   0      0
  0  1    1   0      1
  1  0    1   0      1
  1  1    0   1      1

The text in Boolean algebra#Values states that addition (x+y) is XOR and multiplication (xy) is AND. However, just below that section "Basic operations" uses + and − in the ordinary arithmetic sense where 1+1 is 2. Using those different operators it gives the rule that disjunction (OR) is x + y − xy. Bit confusing. Johnuniq (talk) 07:57, 7 May 2018 (UTC)Reply[reply]

I just added some text to clarify what operators are intended. It's a bit clumsy but better than leaving it as a puzzle. Johnuniq (talk) 02:55, 8 May 2018 (UTC)Reply[reply]

Maybe, but you got it wrong. When truth values are represented as numbers in GF2, they use standard GF2 arithmetic to represent the operations, not logical operations described purely in logic but with the values translated into numbers. Laurent's edit that you reverted was correct but unhelpful, because in GF2 addition and subtraction are the same thing as each other. —David Eppstein (talk) 04:02, 8 May 2018 (UTC)Reply[reply]

Thanks for the fix. For some reason I got the last case (x=1, y=1) confused but I see that mod-2 addition works as you say, and obviously as it has to. Johnuniq (talk) 04:52, 8 May 2018 (UTC)Reply[reply]

This article in many places makes assumption that the underlying set of elements contains only zero and one. But this is not true for boolean algebra as defined in algebra, logic, nor in the historical references that are cited upfront in this article. How can we remedy this confusion? Can we maybe factor the relevant parts that make two-element assumption to Two-element Boolean algebra? Vkuncak (talk) 12:57, 29 May 2021 (UTC)Reply[reply]

According to a paper by Stanley Burris ( https://www.math.uwaterloo.ca/~snburris/htdocs/MYWORKS/PREPRINTS/aboole.pdf): "Contrary to popular belief Boole did not work with a two-element Boolean algebra, nor with the Boolean algebra of subsets of a given set. Boole was simply not doing Boolean algebra, nor Boolean rings."

Tashiro~enwiki (talk) 05:56, 26 December 2021 (UTC)Reply[reply]

See the current status of my new section on Boolean algebra at modus ponens. For discussion, here's my addition just added:

The equivalent concept as formulated in Boolean algebra, the material conditional, expressing x implies y, is notated and defined by ${\textstyle x\rightarrow y=\neg {x}\vee y}$.

Modus ponens and modus tollens follow from the definition above, the rules of negation, and the commutative constant identity for disjunction:

${\displaystyle {\begin{aligned}\neg {0}&=1\\\neg {1}&=0\\\neg {(\neg {x})}&=x\\(\neg {x}=y)&\rightarrow (x=\neg {y})\\x\vee 0&=x\\0\vee x&=x\\\end{aligned))}$

For modus ponens, assuming ${\displaystyle x\rightarrow {y}=1}$ (x implies y is true) and ${\displaystyle x=1,}$ we have:

${\textstyle 1=x\rightarrow y=1\rightarrow y=\neg {1}\vee y=0\vee y=y=1}$

For modus tollens, again assuming ${\displaystyle x\rightarrow {y}=1}$ and ${\displaystyle y=0}$, we have:

${\textstyle 1=x\rightarrow y=x\rightarrow 0=\neg {x}\vee 0=\neg {x))$

${\textstyle \neg {1}=\neg (\neg {x})=x=0}$

This is Boolean algebra as filtered through how an ancient CS guy conceptualizes symbolic reasoning. Pedantic we can do. My problem is to make the rule set complete, I seemed to need to add a problematic rule:

${\displaystyle (\neg {x}=y)\rightarrow (x=\neg {y})}$

This is problematic because it seems to hoist a level above the definition of rightarrow recently introduced.

I scanned the rules here at this page, and my extremely formal and pedantic justification for doing this seems lacking. I'm not even sure this page has defined equality as an operator resolving to a truth value (as common in many computer languages). A better mathematician than I will spot whether this is indeed an axiomatic gap at this page, or a merely an expository gap, or merely a gap of implied application, or even more merely the braino of an antique from across the aisle.

Also, if you can standardize the pedagogy, fill your boots. I was trying to avoid the arithmetic metaphor for this slight application, though perhaps that was misguided.

Note that I'm an extreme tumbleweed editor and I'm unlikely to return here; this is already far more time than I ordinarily expend in any one place. — MaxEnt 02:22, 14 October 2022 (UTC)Reply[reply]

Just to be clear, my strange rule reads to a CS guy as "rewrites as" at the level of whole equations. I'm well aware this is a frame jump, even if it also happens to be algebraically sound. — MaxEnt 02:27, 14 October 2022 (UTC)Reply[reply]

You seem to be looking for the law of the excluded middle, but this is a distinct axiom, and if it is rejected, then you get intuitionistic logic. From there, things snowball into Heyting algebras and assorted other logics. Much of it is not straight-forward. 67.198.37.16 (talk) 01:48, 28 April 2023 (UTC)Reply[reply]

The word "countable" does not appear in this article. Nor is it mentioned that some boolean algebras are not power sets. Never mind that there are (40-year-old) classification theorems for countable boolean algebras. I would love it if this was remedied, either in this article, or in some other. Alas, I am not finding such an exposition in Wikipedia. I do not have the wherewithal to be bold and write such content myself. Anyone? 67.198.37.16 (talk) 01:40, 28 April 2023 (UTC)Reply[reply]

Oooh, seems that Boolean algebras canonically defined begins to tackle this. Still, this article should mention this. 67.198.37.16 (talk) 02:05, 28 April 2023 (UTC)Reply[reply]

This article is not about Boolean algebras — that's at Boolean algebra (structure). This is about "Boolean algebra" as a mass noun — basically the same as the propositional calculus. --Trovatore (talk) 02:12, 28 April 2023 (UTC)Reply[reply]