I'm taking an introductory course in boolean algebra, and have been assigned the task of proving DeMorgan's Laws (so, disclaimer, this is homework). One line of reasoning I came up with is the following:
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By applying the duality principle to two valued boolean algebra, $X \cdot Y = 1 \Leftrightarrow X + Y = 0$
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$X+Y = 0 \Leftrightarrow \overline{X + Y} = 1$
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Therefore, $X \cdot Y = 1 \Leftrightarrow \overline{X + Y} = 0$
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By transitivity of equality, $X \cdot Y = \overline{X + Y}$
where $\cdot$ and $+$ are the conjunction and disjunction operators respectively.
EDIT: As pointed out by Lord_Farin, this result is incorrect, since the conclusion conflicts with DeMorgan's law. Where am I going wrong?
Rest of original question:
Now, the part I'm unsure about is the last step. Have I only managed to prove 4 in the specific case where the claim $X \cdot Y = 1$ is true, or is the proof valid regardless of whether $X \cdot Y = 0$? To my understanding, I have only made claims about the implications of the statement, $X \cdot Y = 1$ but haven't actually claimed whether it is true or false.
Is there a flaw in this proof?
Best Answer
Your error results from misquoting the duality theorem. Let me state it for you here:
But "$X\cdot Y =1$" is certainly not a theorem of BAs; it's contingent (it may be either true or false).
So while you can use duality to conclude the other part of De Morgan's laws:
when you've proven one of them (and in fact, duality is an elegant and efficient method to do it), it cannot be used to prove both of them.