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One thing I found interesting was that Boole’s original algebra was based on AND (×) and XOR (+, exclusive disjunction). In the finite field F2 (i.e. integer ar

One thing I found interesting was that Boole’s original algebra was based on AND (×) and XOR (+, exclusive disjunction). In the finite field F2 (i.e. integer arithmetic mod 2), these are multiplication and addition, respectively. It was later people who rewrote results using AND (∧) and OR (∨, inclusive disjunction) and named this (slightly different) algebra "Boolean algebra". While this has some cute symmetries, the version with XOR is in many ways easier to work with, since it is more compatible with people’s standard expectations for arithmetic (using × and +) on rational numbers. NOT x is also very easy to express: It is just 1 – x (or alternately 1 + x, since x = –x in F2).

082349872349872 4y ago
Are you sure? cf eqn 31 (or p178 of http://www.paradise.caltech.edu/ist4/lectures/Hailperin%20Boole%27s%20Algebra%20isnt%20Boolean%20Algebra.pdf http://www.paradise.caltech.edu/ist4/lectures/Hailperin%20Bo... : Thus by restricting himself to the idempotents of his algebra, Boole would have had Boolean algebra- and we rightfully honor him by attaching his name to this algebra. Only Boole didn't know it. He steadfastly refused to acknowledge any operation but his +. When Jevons claimed that x + x = x, and Boole emphatically denied this, they were really talking about different operations.)
Chinjut 4y ago
Boole didn't work with XOR, as such, and certainly not with F2. Boole worked with polynomials over the integers. Not all such polynomials directly represented propositions for him. But those which sent 0 or 1 valued inputs to 0 or 1 valued outputs did represent propositional formulae (with 0 as false and 1 as true). Thus, he used a notion of + for which, in the familiar way, 1 + 1 = 2, distinct from 0. And then he was able to interpret the inclusive disjunction of A and B as 1 - (1 - A)(1 - B) = A + B - AB. But most often, he spoke of this in cases where A and B were already known to be disjoint from each other, so that AB = 0 and this simplifies to A + B. That presumed disjointness/exclusiveness of A and B might be confused with this being XOR. But he did not use the principle A + A = 0. He worked firmly within familiar non-modular algebra. (He spoke of all this in a fashion a bit less familiar to modern eyes than the way I'm describing it, or anyone else in this discussion is. He never talks about "the OR operation" or any such thing, for example. He just saw himself as applying ordinary numeric algebra to the laws of thought/logic, not making up some new algebraic structure. But the above is indeed in essence what he did.)
- jacobolus 4y ago
  Okay you are right, fair enough. But his equations are consistent with working over F2 (where we can then throw in additional simplifications) and treating multiplication as AND and addition as XOR, in a way that using an algebra of ∧ and ∨ is not. When Boole writes "x xor y" as x(1 − y) + (1 − x)y = 1 ⟺ x − 2xy + y = 1 (p. 53, eq. 31), we can simplify by 2 = 0 and write x + y = 1, treating + here to mean exclusive or.

082349872349872 4y ago

Are you sure? cf eqn 31 (or p178 of http://www.paradise.caltech.edu/ist4/lectures/Hailperin%20Boole%27s%20Algebra%20isnt%20Boolean%20Algebra.pdf http://www.paradise.caltech.edu/ist4/lectures/Hailperin%20Bo... : Thus by restricting himself to the idempotents of his algebra, Boole would have had Boolean algebra- and we rightfully honor him by attaching his name to this algebra. Only Boole didn't know it. He steadfastly refused to acknowledge any operation but his +. When Jevons claimed that x + x = x, and Boole emphatically denied this, they were really talking about different operations.)

Chinjut 4y ago

Boole didn't work with XOR, as such, and certainly not with F2. Boole worked with polynomials over the integers. Not all such polynomials directly represented propositions for him. But those which sent 0 or 1 valued inputs to 0 or 1 valued outputs did represent propositional formulae (with 0 as false and 1 as true). Thus, he used a notion of + for which, in the familiar way, 1 + 1 = 2, distinct from 0. And then he was able to interpret the inclusive disjunction of A and B as 1 - (1 - A)(1 - B) = A + B - AB. But most often, he spoke of this in cases where A and B were already known to be disjoint from each other, so that AB = 0 and this simplifies to A + B. That presumed disjointness/exclusiveness of A and B might be confused with this being XOR. But he did not use the principle A + A = 0. He worked firmly within familiar non-modular algebra. (He spoke of all this in a fashion a bit less familiar to modern eyes than the way I'm describing it, or anyone else in this discussion is. He never talks about "the OR operation" or any such thing, for example. He just saw himself as applying ordinary numeric algebra to the laws of thought/logic, not making up some new algebraic structure. But the above is indeed in essence what he did.)

jacobolus 4y ago
Okay you are right, fair enough. But his equations are consistent with working over F2 (where we can then throw in additional simplifications) and treating multiplication as AND and addition as XOR, in a way that using an algebra of ∧ and ∨ is not. When Boole writes "x xor y" as x(1 − y) + (1 − x)y = 1 ⟺ x − 2xy + y = 1 (p. 53, eq. 31), we can simplify by 2 = 0 and write x + y = 1, treating + here to mean exclusive or.