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- Jan 26, 2012

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$$

\begin{array}{c|c|c|c|c|c}

x &y &z &x \uparrow y &x \oplus z &(x \uparrow y) \oplus (x \oplus z) \\ \hline

0 &0 &0 &1 &0 &1 \\

0 &0 &1 &1 &1 &0 \\

0 &1 &0 &1 &0 &1 \\

0 &1 &1 &1 &1 &0 \\

1 &0 &0 &1 &1 &0 \\

1 &0 &1 &1 &0 &1 \\

1 &1 &0 &0 &1 &1 \\

1 &1 &1 &0 &0 &0

\end{array}

$$

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- Jan 26, 2012

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Actually not. The NAND operation means "not both". If I say $x \uparrow y$, or $x$ NAND $y$, that is equivalent to $\overline{xy}$. By DeMorgan, $ \overline{xy}= \bar{x}+ \bar{y}$, not $ \underbrace{\overline{xy}= \bar{x} \bar{y}}_{\text{Wrong!}}$.Okay, I see where I went wrong. But you're saying x NAND y in the first row is0 NAND 0which is really1 AND 1= 1.

Again, this reasoning is flawed. If you need to, calculate NAND's like this: to compute $x \uparrow y$, first compute $xy$, and then negate the result. You cannot compute NAND's by negating the $x$ and $y$ first, and then AND'ing the results.I get that. Then you are saying in the 2nd row the same thing 0 NAND 0 = 1 because it's really 1 AND 1. Then in the 3rd row you are saying 0 NAND 1 which is really 1 AND 0 = 1. But how does that = 1? 1 AND 0 = 0. It only equals 1 if both are 1.

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