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I am not sure if this Boolean system is correct in both reverse and forward direction ?

$\begin{gathered} \left\{ {\begin{array}{*{20}{l}} {a + b + c}& \equiv &1&{{\text{mod}}\;2}&{\;(1)} \\ {b + c + d}& \equiv &1&{{\text{mod}}\;2}&{\;(2)} \end{array}} \right.\; \hfill \\ \Rightarrow \;a + d \equiv 0\;{\text{mod}}\;2\;\;(3) \hfill \\ \end{gathered} $

Tuong Nguyen Minh
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  • Hey Toung Nguyen Minh, could you please explain the notation to us? You have both $\iff$, and $=$ in the same equation. In the context of Boolean algebra, normally only one of $=$ or $\iff$ is used. Also, could you please explain the meaning of placing an $\iff$ between 2 equations instead of 1? – user400188 Aug 21 '20 at 12:07
  • @JeanMarie in some engineering courses on Boolean algebra, $=$ is used as an iff. In my course, setting something equal to one would have left the expression unaffected, and setting it to zero would negate it. – user400188 Aug 21 '20 at 12:10
  • @user400188 the $\iff$ sign has to be taken in its mathematical (non boolean logic) meaning. – Jean Marie Aug 21 '20 at 12:10
  • @JeanMarie Why is that? Also, should we move this to chat (https://chat.stackexchange.com/rooms/111815/discussion-between-user400188-and-anindya-prithvi)? Edit: Oh I see you changed it to $\iff$. – user400188 Aug 21 '20 at 12:12
  • @user400188 I think the OP means "saying that $A \oplus B \oplus C$ is true and $B \oplus C \oplus D$ is true is equivalent to say that $A \oplus D$ is false. Could the OP confirm this interpretation ? (with $\oplus = $ xor operation). – Jean Marie Aug 21 '20 at 12:16
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    @JeanMarie Ah yes I am confirming this is true – Tuong Nguyen Minh Aug 21 '20 at 13:08
  • You should have kept your initial formulation with $A \oplus B \oplus C=1$ etc. which was perfectly fine. – Jean Marie Aug 21 '20 at 21:15

1 Answers1

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In fact, the $\implies$ direction is true. Not the other one.

Indeed, converting your expressions into their equivalent in modulo 2 arithmetic, it is true that :

$$\begin{cases}a+b+c &\equiv& 1 & \text{mod} \ 2 & \ (1)\\b+c+d &\equiv& 1& \text{mod} \ 2 & \ (2)\end{cases} \ \implies \ a+d \equiv 0 \ \text{mod} \ 2 \ \ (3)$$

Just add for that (1) and (2), using the rule $2x \equiv 0 \ \text{mod} \ 2$, to get (3).

In the reverse direction: taking $a=d=0$ and $b=c=1$, equation (3) is fulfilled while neither equation (1) nor equation (2) is fulfilled.

Jean Marie
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