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Here are my steps: $\bar{x}\bar{y}\bar{z} + \bar{x}\bar{y}z+x\bar{y}\bar{z}+x\bar{y}z$ $$\bar{x}\bar{y}\bar{z} + \bar{x}\bar{y}z+x\bar{y}\bar{z}+(x\bar{y}z+x\bar{y}z)$$ Used Idempotent law and rearranged the equation $$\bar{x}\bar{y}\bar{z} + (\bar{x}\bar{y}z+x\bar{y}z)+(x\bar{y}\bar{z}+x\bar{y}z)$$ factoring $$\bar{x}\bar{y}\bar{z} + \bar{y}z(\bar{x}+x)+x\bar{y}(\bar{z}+z)$$Complement law $$\bar{x}\bar{y}\bar{z} + \bar{y}z(1)+x\bar{y}(1) + \bar{x}\bar{y}\bar{z} $$ Used Idempotont law again $$\bar{x}\bar{y}\bar{z} + \bar{y}(x + \bar{x}+z+\bar{z}) $$factoring $$\bar{x}\bar{y}\bar{z} + \bar{y}(1+1) $$Complement Law and Tautology $$\bar{x}\bar{y}\bar{z} + \bar{y}1$$Used Identity law on y $$\bar{y}(1+\bar{x}\bar{z})$$Tautology $$\bar{y}$$ However I need x and z values. $$\bar{y}+\bar{x}x+z\bar{z}$$ The question states that I need to find the minimization of the original expression, as the sum of three terms. Is my steps valid?

Dravid
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    Hard to see what you did from step to step. Especially the last step when you suddenly arrive at just "y-bar". Also: what is the line right after "However I need..."? Is it the answer you're supposed to get to, i.e. your problem is an exercise with that line given as the answer? – coffeemath Dec 17 '22 at 11:02
  • @coffeemath The question states that I need to find the minimization of the original expression, as the sum of three terms. Sorry if the steps are unclear, I will add some comments next to them now – Dravid Dec 17 '22 at 11:17
  • Good to insert the rule used for each step. Also there can be no valid rule to arrive at y-bar. The starting expression is not equivalent to y-bar. – coffeemath Dec 17 '22 at 11:27
  • You still did not say whether the 3-term expression of the last display is the book answer, or is just one of your steps. – coffeemath Dec 17 '22 at 11:28
  • @coffeemath The final expression is my answer for the question. Sadly the book doesnt have an answer key – Dravid Dec 17 '22 at 11:30
  • What are $x,y,z $? What is the bar operation? – P. Lawrence Dec 17 '22 at 12:27
  • @P.Lawrence x, y, and z are boolean variables and the bar operation stands for complement/negation – Dravid Dec 17 '22 at 12:28
  • Let me use $x'$ for $\bar{x}$, as it is easier to typeset. There is a point in which you have the expression $$x'y'z'+y'z(x+x')+xy'(z+z')$$ which simplifies to $x'y'z'+y'z+xy'$, but then you add another $x'y'z'$, and claim that by idempotency you get another expression which I really can't see how. But $$x'y'z'+y'z+xy'=y'(x+x'z'+z)$$ and $$x+x'z'+z=x(z+z')+x'z'+z=xz+z+(x+x')z'=xz+z+z'=1.$$ Hence the original expression simplifies to $y'$ (I just can't match your steps to the justifications.) – amrsa Dec 17 '22 at 12:41
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    @amrsa I will try simplifying it again.

    I also tried finding the simplification using a karnaugh map, and reached the same answer = y'

    But the truth table for the original expression and y' dont match, so I cannot present that as the final answer https://math.stackexchange.com/questions/4600582/karnaugh-map-for-the-expression-barx-bary-barz-barx-baryzx-bary

     – Dravid Dec 17 '22 at 12:57
  • @Dravid They must match. Or else we did something wrong! – amrsa Dec 17 '22 at 13:01
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    Note that the expression is simply $$y' (xz + xz' + x'z + x'z') = y' (x + x') (z + z') = y'.$$ – L. F. Dec 18 '22 at 00:31

1 Answers1

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  1. Your simplification process is correct, and so is your result $\bar y$.
  2. If your truth tables don't match then you probably made a mistake computing the truth tables. I can't check those because you haven't posted them.
  3. Your last statement "The question states that I need to find the minimization of the original expression, as the sum of three terms." is very strange since the expression simplifies to just 1 term. I realize you can awkwardly rewrite it as 3, but I would expect whoever assigned this question would not ask for 3 terms if the expression simplifies to just $\bar y$. It makes me think you should double-check for mistakes in copying down the initial expression to be simplified. (That could also explain why your truth tables aren't matching up.)
David Clyde
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  • I just realized I might be inputting the truth tables incorrectly. For the initial expression, the truth table has 3 boolean variables ( 8 rows), and the final simplification $\bar{y}$ has just 1 variable (2 rows). Do I have to include x and z values to the $\bar{y}$ truth table too? – Dravid Dec 18 '22 at 05:09
  • @Dravid You've shown the original expression is equivalent to the three-variable truth table $\bar y$. We're viewing $\bar y$ as a function of 3 variables, so it will produce a 2x2x2 truth table that just doesn't depend on $x$ or $z$ at all. – David Clyde Dec 18 '22 at 05:18
  • I don't understand how $\bar{y}$ can have 2x2x2 truth table. Would the columns be like $y$, $\bar{y}$? Sorry if I misunderstood – Dravid Dec 18 '22 at 05:25
  • @Dravid $x$ could be true or false. $y$ could be true or false. $z$ could be true or false. Therefore $f(x,y,z) = \bar y$ has 8 different inputs which need to be shown in that table, even though the output doesn't depend on $x$ or $z$. – David Clyde Dec 18 '22 at 05:28
  • That really helps, thank you so much David! – Dravid Dec 18 '22 at 05:36