gill1109 wrote:I would use
Z: AB (AB == 1) ,
having explained that in an arithmetical expression, logical expressions are converted from T/F to 1/0. I would use == when asking are LHS and RHS equal, forcing AB == 1 to be read as logic. I would use := for defining LHS by RHS, and plain = for a chain of expressions asserting equality.
In Z, three functions are multiplied together: A, B, and (AB == 1). The third function is a function of a, b and lambda and it takes values in {0, 1}.
A(a, lambda) is a number
A is a function
You did not define “x is equivalent to 1”, and the set of numbers x equal to 1 contains only one number.
So X and Y are awful. Say simple things in a simple way. Mathematical notation is designed in order to say difficult things as clearly and simply as possible. Mathematics is built up logically and sequentially. Every time you introduce new notation, or you change notation, you should say so. Usually you are writing in a context which means that you know the notation which your readers are familiar with, and you use it too, if possible, or at least, to start with. You build on foundations which are there and which are known by your readers.
I realise how difficult it must be! I have had these discussions with students for 45 years and once I was a student myself. I remember a student asking me (after the lecture) “does the less than or equals symbol” mean less than or equals, or less than and equals? I wish he had dared to ask it during the lecture.
Thanks Richard,
I need to see your proposal in proper math notation.
Note that I was using a formal set-builder notation (text-book stuff). I now think it clearer if I make conditional statements.
SO: This next is very clear and rigorous to me. How about you?
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This is Bell’s theorem, BT: E(a,b|β) = ∫dλ ρ(λ)A(a,λ)B(b,λ) ≠ −a·b [sic]; (1)
with A(a,λ) = ±1 ≡ A^±, B(b,λ) = ±1 ≡ B^±, A(a,λ)B(b,λ) = ±1 ≡ AB in short form. (2)
REFUTATION: Via RHS (2), and independent of the functions A and B, we distribute (1)’s integrand
(identified in short form as AB = ±1) over two subsidiary integrands: AB = 1 and AB = −1.
Thus:
E(a,b|β) = ∫dλ ρ(λ)[(A(a,λ)B(b,λ)|AB = 1)−(A(a,λ)B(b,λ)|AB = −1)] (3)
= P (AB = 1) − P (AB = −1) , the weighted-sum of the binary AB results ±1. (4)
etc. etc. Gordon