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Plz Solve this Boolean Expression using Boolean Algebra & Plz Mention the Laws used to Simplify it. Thanks.

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closed as off-topic by abiessu, Graham Kemp, iadvd, user91500, JonMark Perry Aug 23 '16 at 5:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – abiessu, Graham Kemp, iadvd, JonMark Perry
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Please show us what you have tried. $\endgroup$ – Graham Kemp Aug 23 '16 at 4:44
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    $\begingroup$ Please use proper spelling to the best of your abilities. $\endgroup$ – Em. Aug 23 '16 at 4:47
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Just writing out the 8 lines truth table would be revealing.

Alternatively: \begin{eqnarray} XY+\overline{XZ}+X\overline{Y} Z &=& XY+\overline{X}+\overline{Z}+X\overline{Y} Z \\ &=& XY+\overline{X}+\overline{X}Y+\overline{Z}+X\overline{Y} Z \\ &=& Y+\overline{X}+\overline{Z}+X\overline{Y} Z \\ &=& Y+\overline{X}+\overline{Z}+X\overline{Y} \overline{Z} +X\overline{Y} Z \\ &=& Y+\overline{X}+\overline{Z}+X\overline{Y} \\ &=& Y+\overline{X}+\overline{X}\overline{Y}+\overline{Z}+X\overline{Y} \\ &=& Y+\overline{X}+\overline{Z}+\overline{Y} \\ &=& 1 \end{eqnarray}

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  • $\begingroup$ Thank you. Are there any good websites for learning Boolean Algebra ? $\endgroup$ – Sandy Action Aug 23 '16 at 4:56
  • $\begingroup$ I didn't understand the second line. Please explain. $\endgroup$ – Sandy Action Aug 23 '16 at 5:06
  • $\begingroup$ It is always true that $A = A+AB$. $\endgroup$ – copper.hat Aug 23 '16 at 5:07
  • $\begingroup$ Yes, I understood. A+AB=A(1+B)=A(1)=A $\endgroup$ – Sandy Action Aug 23 '16 at 5:18
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    $\begingroup$ I don't understand what you are asking. It is always the case that $A+\overline{A} = 1$. $\endgroup$ – copper.hat Aug 23 '16 at 5:30
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Begin with deMorgan's negation

$\rm \quad XY+(XZ)'+ X Y'Z \\ = XY+(X'+ Z')+ X Y'Z$

From there use distribution and complementation until you can't use it any more.

That is: $\rm\quad A+A'B \\ = (A+A')(A+B)\\ = A+B$

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  • $\begingroup$ My question was X Y + (X Z)' + X Y' Z.......not XY+(XZ)′+Y′Z $\endgroup$ – Sandy Action Aug 23 '16 at 5:05
  • $\begingroup$ Copy error, though it makes little difference. $\endgroup$ – Graham Kemp Aug 23 '16 at 8:00
  • $\begingroup$ ab+(ab)'+a'b .. ab+a'b+a'+b' .. b(a+a')+a'+b'.........b+a'+b'..........b+b'+a'.......1+a'...‌​.= 1 This is what I did. It is coming 1 where as if I do by your process it is coming A+B. Which one is correct ? $\endgroup$ – Sandy Action Aug 23 '16 at 8:04
  • $\begingroup$ @SandyAction What does that have to do with the problem asked? $\endgroup$ – Graham Kemp Aug 23 '16 at 8:09
  • $\begingroup$ No, I am asking for the simplified form... Whats it ? 1 or X+Y ? Sorry, I had changed the variables in my previous comment ... A for X, B for Y & C for Z. $\endgroup$ – Sandy Action Aug 23 '16 at 8:14

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