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Simplify the expression: $((XY+X')\oplus (Y+YZ))'$

I can go as far as: $$((XY+X')\oplus (Y+YZ))'$$ $$\text{I use the rule of absorption:}$$ $$((Y+X')\oplus Y)'$$ Now I see that if $Y=1$ than the result is one and if $Y=0$ the result is $X$, but according to Wolfram the result is $X+Y$.

Where did I go wrong?

2 Answers 2

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We know that: $$ A \oplus B = \bar A B + A \bar B$$ and that $\overline {A\cdot B} = \bar A+ \bar B $ and $\overline {A+B} = \bar A \bar B $. Thus, $$((Y + \bar X) \oplus Y) = \overline {Y + \bar X} Y + (Y + \bar X) \bar Y = (X\bar Y Y) + (Y\bar Y + \bar X \bar Y) = 0 + 0 + \bar X \bar Y $$

Thus, $$\overline {\bar X \bar Y} = X + Y $$ Hope it helps.

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    How did you get from $\overline {Y + \bar X} Y + (Y + \bar X) \bar Y$ to $(X\bar Y Y) + (Y\bar Y + \bar X \bar Y)$? – 2017-02-16
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    @PanthersFan92 For the first term, remember that $\overline{X+Y}=\overline{X}\overline{Y}$ and $\overline{\overline{X}}=X$. So $\overline{Y+\overline{X}}Y=\overline{Y}\overline{\overline{X}}Y=\overline{Y}XY$. For the second term, use distributivity: $(Y+\overline{X})\overline{Y}=Y\overline{Y}+\overline{X}\overline{Y}$. – 2017-02-16
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You didn't go wrong!

Notice that in general $X \oplus (X + Y) \Leftrightarrow X'Y$

(since $X$ cannot be true, as that would make both sides true ... and hence $Y$ should be true)

So, where you left off:

$$((Y+X')\oplus Y)' \Leftrightarrow$$

$$(X'Y')' \Leftrightarrow$$

$$X + Y$$

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    How come? $(X'Y)' \Leftrightarrow X + Y$? – 2017-02-16
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    @PanthersFan92 Oops, typo on my end. It should be $(X'Y')'$ on the second line. Fixed in answer now – 2017-02-16