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How can I prove that:

$$ ((a + b)\bmod{2} \; + \;c)\bmod{2} = (a+b+c)\bmod{2} $$

I am unsure of the name of this property, but I realized I know this by intuition. The reason for this is that I want to take a sum over $\mathbb{F_2}$ and I would like to say that

$$ (\sum_{i}^{} a_i)\bmod{2} $$

will suffice, as opposed to taking individual modulus of every addition operation.

Bill Dubuque
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user129393192
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  • My question is not answered by the linked posts. I would like a proof of the above. – user129393192 May 12 '23 at 17:21
  • As explained here in the linked dupes, by the Congruence Sum & Product rules, we obtain a congruent expression by replacing aguments of sums & products by any congruent argument , e.g. $,n,\to n\bmod 2,$ or reversely). – Bill Dubuque May 12 '23 at 17:45
  • As explained there, your $!\bmod 2$ equality is equivalent to $,\color{#c00}{(a+b)\bmod 2}+ c\equiv a+b+c\pmod{!2},$ which follows by replacing the summand $,\color{#c00}{(a+b)\bmod 2},$ by $,\color{#c00}{a+b},,$ which is valid since, again, they are congruent $!\pmod{!2}\ \ $ – Bill Dubuque May 12 '23 at 17:51

1 Answers1

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You can instead prove it on two variables first that, $$(a+b(\mod 2))\mod 2) \equiv a(\mod2)+b(\mod2)$$

After that you can assume $z=a+b$ and reuse your proof on $z$ and $c$. This holds true for any number of variables.

Learningstill
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