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I start with: $$\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}\bar{C}+A\bar{B}{C}=A\bar{B}+\bar{B}\bar{C}+\bar{A}BC$$ then I did: $$\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}\bar{C}+A\bar{B}{C}=\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}(\bar{C}+C)=\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}$$

How can I show that: $$\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}=A\bar{B}+\bar{B}\bar{C}+\bar{A}BC$$ I tried taking out $$\bar{A}(BC+\bar{B}\bar{C})+A\bar{B}$$ but then I get stuck since whatever I try leads me deeper into the forest of boolean algebra.

Edit: enter image description here

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  • $\begingroup$ I don't think that's true... $\endgroup$ – user142299 Apr 19 '14 at 2:44
  • $\begingroup$ @NotNotLogical I got this from Digital Design and Computer Architecture, the example above is a part of the truthtable that I am having a problem with since the other part cancels I'll edit the problem to make the solution clearer. $\endgroup$ – Vector_13 Apr 19 '14 at 3:02
  • $\begingroup$ It's not a true equivalence claim. The left side must have $A$ be false, so the left side cannot equal the right side. $\endgroup$ – user142299 Apr 19 '14 at 3:04
  • $\begingroup$ @NotNotLogical Yes, you are right I didn't write the problem correctly, I'm sorry, I will edit the post. Thanks. $\endgroup$ – Vector_13 Apr 19 '14 at 3:16
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$$\bar{A}\bar{B}\bar{C}+\bar{A}BC+A\bar{B}\bar{C}+A\bar{B}{C}=A\bar{B}+\bar{B}\bar{C}+\bar{A}BC$$

I'm going to use lower-case letters and $a'$ for $\bar{a}$ if you don't mind. First subtract (!!) the expression $$a'bc$$ from both sides to get $$a'b'c'+ab'c'+ab'c=ab'+b'c'$$ Then factor on the left, $$b'c'(a'+a)+ab'c=ab'+b'c'$$ Simplify $$b'c'+ab'c=ab'+b'c'$$ Factor again $$b'(ac+c')=ab'+b'c'$$ and again $$b'(c'+c)(c'+a)=ab'+b'c'$$ $$b'(c'+a)=ab'+b'c'$$ which is certainly true.

Word of advice - don't do that "subtract" thing ever, because it does not necessarily yield equivalent expressions. I simply discarded one piece because it was not needed.

Also, in the second-to-last line I used the identity $$x+yz=(x+y)(x+z)$$

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  • $\begingroup$ Wow, that is beautiful, thanks so much. $\endgroup$ – Vector_13 Apr 19 '14 at 3:46

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