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The expression is: F = (X' + Y' + Z')(Y' + A')

I have no clear idea on how to go about simplifying this with Boolean algebra. After it's simplified, I'll need to implement it only using NAND gates.

My instinct is to start off with using the distributive property? Or would it be easier to use k-maps?

borderlineNovice
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2 Answers2

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First of all, your question is wrong place however your answer is,

$1. \quad F = (X' + Y' + Z')(Y' + A')$

$2. \quad F=(X'Y'+X'A+Y'+Y'A+Z'Y'+Z'A)$

$3. \quad F=( Y'(X'+Z') + A(X'+Z') + Y'(A+1) )$

$4. \quad F=( Y'(X'+Z'+1) + A(X'+Z') )$

$5. \quad F=( Y'(1) + A(X'+Z') )$

$6. \quad F=( Y' + AX' + AZ' )$

Well, How can I draw only NAND gates diagram? Firsty, I have to know that $(AA)'$ is $A'$ and I can obtain not of anything in these question, then you can step back in circuit for correct usage of NANDs.

click here to access circuit

daOnlyBG
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mathema
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You start with your expression and turn it into NANDs, bit by bit. An important rule is that NAND (x, y) = OR (x', y').

F = AND (X' + Y' + Z', Y' + A')

F = NAND (F1, F1) where F1 = NAND (X' + Y' + Z', Y' + A').

F1 = NAND (F2, F3) where F2 = X' + Y' + Z', F3 = Y' + A'.

F2 = NAND (NOT (X' + Y'), Z), so F2 = NAND (F4, Z) where F4 = NOT (X' + Y')

F3 = NAND (Y, A)

F4 = NAND (F5, F5) where F5 = X' + Y'

F5 = NAND (X, Y)

gnasher729
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