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I want to know, how to prove this,

$(AC + BD)(AC)'(BD)' = (AC)(AC)' + (BD)(BD)'.$

Please, help me.... Thanks. What laws in that line?

  • We know that $xx' = 0$. Hence $(AC)(AC)' = 0$ and $(BD)(BD)' = 0$. Therefore the whole expresion equals $0$. Why do you need to simplify an expression which is equal to $0$? – Anastassis Kapetanakis Apr 05 '20 at 16:31
  • Cuz, To prove it. I want to prove (AC+BD)(AC+BD)'=0. –  Apr 05 '20 at 16:35

1 Answers1

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So what you are trying(as I understand from your comment) is to prove that $(AC + BD)(AC)'(BD)' = 0$. We have: $$(AC + BD)(AC)'(BD)' = (AC)(AC)'(BD)'+ (BD)(AC)'(BD)' = (AC)(AC)'(BD)' + (BD)(BD)'(AC)'$$

We know that $xx' = 0$. So $(AC)(AC)' = 0$ and $(BD)(BD)' = 0$. Therefore: $$(AC + BD)(AC)'(BD)' = (AC)(AC)'(BD)' + (BD)(BD)'(AC)' = 0 + 0 = 0$$

Another proof(by De Morgan's law): $$(AC + BD)(AC)'(BD)' = (AC + BD)(AC + BD)' = 0$$

Anastassis Kapetanakis
  • 1,020
  • Hmmm... Sorry for my Question. and Thank you. But, In Solution. There is one more step. =(AC+BD)(AC)'(BD)' =(AC)(AC)' + (BD)(BD)' =((AC) + (AC)')' + ((BD) + (BD)')' So, I want to know that step. I mean, What laws in that line. In second line. –  Apr 05 '20 at 16:55
  • I've done a typo mistake. I corrected it! – Anastassis Kapetanakis Apr 05 '20 at 17:06
  • I checked it, But, I want to know how to prove my question. I will accept you in few minutes later, But, can u check again? –  Apr 05 '20 at 17:14
  • Actually Full prove is here. = (AC + BD)(AC)'(BD)' = (AC)(AC)' + (BD) (BD)' = ((AC) + (AC)')' + ((BD) + (BD)')' = (1)' + (1)' = 0 I just want to know that line. Thanks –  Apr 05 '20 at 17:16
  • I cannot figure out how one can go for the first line to the second. Maybe it's a mistake. I dont't think that there is a law that allows such an operation. But I provided an alternative proof that is quicker that the first one I showed. I apologise if I could't help you! – Anastassis Kapetanakis Apr 05 '20 at 17:42
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    OK, Thank you. Have a niceday! –  Apr 05 '20 at 17:44