Is it logically valid to prove DeMorgan's laws using the duality of boolean algebra?

Question

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:

1. By applying the duality principle to two valued boolean algebra, $X \cdot Y = 1 \Leftrightarrow X + Y = 0$

2. $X+Y = 0 \Leftrightarrow \overline{X + Y} = 1$

3. Therefore, $X \cdot Y = 1 \Leftrightarrow \overline{X + Y} = 0$

4. 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?

Answer 1

Your error results from misquoting the duality theorem. Let me state it for you here:

If $T$ is a theorem about boolean algebras, then so is $T^*$, the statement obtained by carrying out the replacements $+ \leftrightarrow \cdot$ and $1 \leftrightarrow 0$.

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:

• $\overline X \cdot \overline Y = \overline{X+Y}$
• $\overline X + \overline Y = \overline{X \cdot Y}$

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.