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Okay, so I started from ( (W + Y + Z')' (W' + X' + Y)' (X' + Y' + Z')' (W + Y + Z)' )'

then simplified further to:

( W'Y'Z + WXY' + XYZ + W'Y'Z' )'

then:

(W + Y + Z')(W' + X' + Y)(X' + Y' + Z')(W + Y + Z)

to:

(W + Y)(W' + X' + Y)(X' + Y' + Z')

The answer is: (W + Y)(X' + Y)(X' + Z').

How do I go from what I currently have to the answer? I've tried breaking up the other two expressions to (W' + X')(W' + Y) and played around like that, but to no avail. What rule am I not using correctly?

Thanks

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  • $\begingroup$ This part is wrong: $( W'Y'Z + WXY' + XYZ + W'Y'Z' )'$, it should be like this: $( (W'Y'Z)(WXY')(XYZ)(W'Y'Z'))'$ which is equal to 1 $\endgroup$ – Lee Feb 1 at 2:57
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Starting from (W + Y)(W' + X' + Y)(X' + Y' + Z') (you have already solved till this point)

let Y=A, B=W and C=(W'+X') for next step using (A+B)(A+C)=(A+BC);

=(Y+W(W'+X')) (X' + Y' + Z')

=(Y+WW'+WX')(X' + Y' + Z') //WW'=0

=(Y+WX')(X' + Y' + Z') //simplified by removing WW'=0

=(Y+W)(Y+X')(X' + Y' + Z') // used (A+B)(A+C)=A+BC

=(W+Y)(X'+Y(Y'+Z')) // used (A+B)(A+C)=A+BC

=(W+Y)(X'+YY'+YZ')

=(W+Y)(X'+YZ')

=(W+Y)(X'+Y)(X'+Z')=Answer

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  • $\begingroup$ Jesus, didn't think I was so far from solving it O.O Thanks a bunch for the help! $\endgroup$ – ESM Feb 1 at 3:05
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$((W+Y+Z')'(W'+X'+Y)'(X'+Y'+Z')'(W+Y+Z)')'=(W+Y+Z')+(W'+X'+Y)+(X'+Y'+Z')+(W+Y+Z)=W'+W+X'+Y'+Y+Z'+Z=1$

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You can use:

Generalized Reduction

$(P+Q)(P+Q'+R)=(P+Q)(P+R)$

With that:

$(W+Y)(W'+X'+Y)(X'+Y'+Z')=$

$(W+Y)(X'+Y)(X'+Y'+Z')=$

$(W+Y)(X'+Y)(X'+Z')$

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