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I'm currently learning for my maths exam, and in the part about boolean algebra I came across an exercise that I can't seem to solve. I probably only need the first few steps to get started.

$$ (xyz + uv)(x+\overline{y}+\overline{z}+uv) $$

Usually, if I get into trouble, I can fall back to a truth table or VK-diagram, but that's just too much work for 5 variables.

Thanks in advance!

Simon Verbeke
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2 Answers2

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Multiply the terms in the two brackets. You get: $$ xyz + 0 + 0 + xyzuv + xuv + \overline{y}uv + \overline{z}uv + uv$$ $$ xyz(1+uv) + uv(1+x+\overline{y}+\overline{z})$$ $$ xyz + uv $$

NOTE:

$1 + x = 1$

$1.x = x$

$x.x = x$

$x.\overline{x} = 0$

lsp
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Let's write it like this: $$\begin{align} (xyz + uv)((x+\bar y + \bar z) + uv)&\equiv((xyz)(x+\bar y + \bar z) + (xyz)(uv))+(uv(x + \bar y + \bar z) + uvuv) \\ &\equiv ((xyz)(x+\bar y + \bar z) + xyzuv)+(uv(x + \bar y + \bar z) + uv)\\ &\equiv ((xyz)(x+\bar y + \bar z) + xyzuv)+(uv) \\ &\equiv ((xyzx + xyz\bar y + xyz\bar z) + xyzuv)+(uv) \\ &\equiv ((xyz + 0 + 0) + xyzuv)+(uv) \\ &\equiv (xyz + xyzuv)+(uv) \\ &\equiv (xyz)+(uv) \\ \end{align}$$

From line 1 to line 2, we distributed. From 2 to 3, we noted that $ab + a \equiv a$. From 3 to 4, we distributed again. From 4 to 5, we noted that $a\bar a\equiv 0$ (contradiction), which we remove from 5 to 6. 6 to 7 we note $ab + a \equiv a$.

apnorton
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  • Is $ a\bar a\equiv c $ the same as $ a\overline{a} = 0 $ ? I've never seen the first notation, we use the second. – Simon Verbeke Jun 06 '13 at 12:32
  • @SimonVerbeke In the class I took, we used $c$ to denote a contradiction, and $\tau$ to denote a tautology. We also used the $a\land b$ and $a \lor b $ operators instead of $ab$ and $a+b$, respectively. I'll edit my question to use the $0$ to denote contradiction. – apnorton Jun 06 '13 at 12:35
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    Ok, thank you! Yours is a good solution too, but I find lsp's solution simpler. So I'm going to accept his as the answer. – Simon Verbeke Jun 06 '13 at 12:37