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I know that I have to use the rule where (A+B)' is A'*B' in order to solve it but I don't know how to approach the problem.

I tried messing around with the equation and making the complement and then just adding but it doesn't turn out to be the same thing.

Aryan Gupta
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    Please, explain the 'and *operators. – utobi Sep 09 '22 at 16:16
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    @utobi As OP mentions NAND and NOT gates I suppose it's about reducing boolean expression to be implemented as an electronic circuit for digital signal processing. As such, the apostrophe is negation (NOT), the asterisk is conjunction (AND) and the plus is alternative (OR). The objective is then to transform a given expression in three input variables into a form using NAND only, with NOT X equivalent to NOT(X AND X) equivalent to (X NAND X) = (X*X)'. – CiaPan Sep 09 '22 at 18:10
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    @utobi the ' and * operators are standard notations in boolean algebra. – Jean Marie Sep 09 '22 at 19:42

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Recall that you can make AND ($\land$) and OR ($\lor$) gates out of NAND ($\not\land$) and NOT ($\lnot$):

  • AND: $A \land B = \lnot (A \not\land B)$
  • OR: $A \lor B = (\lnot A) \not\land (\lnot B)$

So just straightfowardly implement your expression with AND, OR, and NOT gates; make the above substitutions for AND and OR; and remove any redundant NOT NOT gate combinations.

Dan
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    Additionally, any remaining NOT can be implemented with NAND, too: NOT X = NOT (X AND X) = X NAND X. – CiaPan Oct 07 '22 at 12:52