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The divisibility by 3 rule can be repeated, as shown by the following theorem: Theorem 2. Let x be an integer, x >= 0. Let x have n digits where n >= 1. Let m be an integer, m >= 1. Then: x is divisible by 3 if and only if cross sum**m (x) is divisible by 3 where m is the number of iterations (1) of the cross sum function. Proof. We apply proof by induction (2). The base case of proof by induction, m = 1, is true by Theorem above. What remains to prove is the induction case of proof by induction, which says that if the theorem holds for any given case m = i, it holds for m = i+1. So, we assume that the theorem holds for m = i. This means that: x is divisible by 3 if and only if cross sum**i (x) is divisible by 3 We note that cross sum**i (x) satisfies: cross sum**i (x) is a positive integer as cross sum**i (x) >= 0. cross sum**i (x) have n' digits where n' >= 1. Then Theorem gives: cross sum**i (x) is divisible by 3 if and only if the cross sum of cross sum**i (x) is divisible by 3 in other words: cross sum**i (x) is divisible by 3 if and only if cross sum**(i+1) (x) is divisible by 3. This gives cross sum**i (x) is divisible by 3 = cross sum**(i+1) (x) is divisible by 3 by reference (3) by noting that either side of the equal sign is a Boolean value We have that the theorem holds for m = i: x is divisible by 3 if and only if cross sum**i (x) is divisible by 3 Therefore: x is divisible by 3 if and only if cross sum**(i+1) (x) is divisible by 3 by replacing "cross sum**i (x) i

The divisibility by 3 rule can be repeated, as shown by the following theorem: Theorem 2. Let x be an integer, x >= 0. Let x have n digits where n >= 1. Let m be an integer, m >= 1. Then: x is divisible by 3 if and only if cross sum**m (x) is divisible by 3 where m is the number of iterations (1) of the cross sum function. Proof. We apply proof by induction (2). The base case of proof by induction, m = 1, is true by Theorem above. What remains to prove is the induction case of proof by induction, which says that if the theorem holds for any given case m = i, it holds for m = i+1. So, we assume that the theorem holds for m = i. This means that: x is divisible by 3 if and only if cross sum**i (x) is divisible by 3 We note that cross sum**i (x) satisfies: cross sum**i (x) is a positive integer as cross sum**i (x) >= 0. cross sum**i (x) have n' digits where n' >= 1. Then Theorem gives: cross sum**i (x) is divisible by 3 if and only if the cross sum of cross sum**i (x) is divisible by 3 in other words: cross sum**i (x) is divisible by 3 if and only if cross sum**(i+1) (x) is divisible by 3. This gives cross sum**i (x) is divisible by 3 = cross sum**(i+1) (x) is divisible by 3 by reference (3) by noting that either side of the equal sign is a Boolean value We have that the theorem holds for m = i: x is divisible by 3 if and only if cross sum**i (x) is divisible by 3 Therefore: x is divisible by 3 if and only if cross sum**(i+1) (x) is divisible by 3 by replacing "cross sum**i (x) i

Theorem: Let A, B be Boolean variables (1). Then: (A B) = (A = B) Proof: We prove this by evaluating the condition in the theorem text and demonstrating that the value is true in all cases. There are three possible combinations of the values of A and B: A = B A = false, B = true A = true, B = false We have: (A B) = (A = B) = ((A => B) and (B => A)) = (A = B) by definition of (2) = ((~A or B) and (~B or A)) = (A = B) by definition of => (2) The case A = B: We have: ((~A or B) and (~B or A)) = (A = B) = (((~B or B) and (~B or B)) = (B = B)) as A = B = ((true and true) = (B = B)) by negation law (2) = (true = (B = B)) by idempotent law (2) = (true = true) by reflexivity law (3) = true by reflexivity law (3) which is what we set out to prove. The case A = false, B = true: We have: ((~A or B) and (~B or A)) = (A = B) = (((~false or true) and (~true or false)) = (false = true)) as A = false, B = true = (((true or true) and (false or false)) = (false = true)) as ~false = true and ~true = false (4) = ((true and false) = (false = true)) as true or true = true and false or false = false (1) = (false = (false = true)) as true and false = false (1) = (false = false) as (false = true) = false as false and true are different values (3) = true by reflexivity law (3) The case A = true, B = false: We have: ((~A or B) and (~B or A)) = (A = B) = (((~true or false) and (~false or true)) = (true = false)) as A = true, B = false = (((false or false) and (true or true)) = (true = false)) as ~true =

Theorem: Let A, B be Boolean variables (1). Then: (A B) = (A = B) Proof: We prove this by evaluating the condition in the theorem text and demonstrating that the value is true in all cases. There are three possible combinations of the values of A and B: A = B A = false, B = true A = true, B = false We have: (A B) = (A = B) = ((A => B) and (B => A)) = (A = B) by definition of (2) = ((~A or B) and (~B or A)) = (A = B) by definition of => (2) The case A = B: We have: ((~A or B) and (~B or A)) = (A = B) = (((~B or B) and (~B or B)) = (B = B)) as A = B = ((true and true) = (B = B)) by negation law (2) = (true = (B = B)) by idempotent law (2) = (true = true) by reflexivity law (3) = true by reflexivity law (3) which is what we set out to prove. The case A = false, B = true: We have: ((~A or B) and (~B or A)) = (A = B) = (((~false or true) and (~true or false)) = (false = true)) as A = false, B = true = (((true or true) and (false or false)) = (false = true)) as ~false = true and ~true = false (4) = ((true and false) = (false = true)) as true or true = true and false or false = false (1) = (false = (false = true)) as true and false = false (1) = (false = false) as (false = true) = false as false and true are different values (3) = true by reflexivity law (3) The case A = true, B = false: We have: ((~A or B) and (~B or A)) = (A = B) = (((~true or false) and (~false or true)) = (true = false)) as A = true, B = false = (((false or false) and (true or true)) = (true = false)) as ~true =

I like using the plate method too, it keeps things easy and stops me from overcomplicating what I eat. I noticed when I focus on making half my plate veggies and mix in good proteins and whole grains, I feel fuller and just more energized in general. Sometimes I still get confused with all these diet trends and everything online though, especially when it comes to which supplements, if any, might actually help. Recently, I started checking out Menalam for some clarity on supplements that fit my diet and routine. It's a cool way to get recommendations that actually make sense for my own lifestyle and not just generic advice.

I like using the plate method too, it keeps things easy and stops me from overcomplicating what I eat. I noticed when I focus on making half my plate veggies and mix in good proteins and whole grains, I feel fuller and just more energized in general. Sometimes I still get confused with all these diet trends and everything online though, especially when it comes to which supplements, if any, might actually help. Recently, I started checking out Menalam for some clarity on supplements that fit my diet and routine. It's a cool way to get recommendations that actually make sense for my own lifestyle and not just generic advice.