# Chap 1 Slide 19: Simplifications of the canonical representation

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## Chap 1 Slide 19: Simplifications of the canonical representation

 Hello ! This is my first post so, first of all, I would like to thank you for making this very interesting course available on internet ! I do not understand how the simplifications of the canonical representation on the slide 19 were made: https://docs.wixstatic.com/ugd/56440f_51f7dbaa5d004686a9576588088458c6.pdfI do not understand how we went from: (Not(x) And Not(y) And Not(z)) Or (Not(x) And y And Not(z)) Or (x And Not(y) And Not(z)) to: (Not(x) And Not(z)) Or (x And Not(y) And Not(z)) and even less to: (Not(x) And Not(z)) Or (Not(y) And Not(z)) But I do understand the 2 last lines: (Not(x) And Not(z)) Or (Not(y) And Not(z)) = Not(z) And (Not(x) Or Not(y)) Thank you !
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## Re: Chap 1 Slide 19: Simplifications of the canonical representation

 This post was updated on . The first level is easy... (X'Y'Z') + (X'YZ') + (XY'Z') In the first two terms, we can factor out a common (X'Z') --> (X'Z')(Y' + Y) + (XY'Z') Y' + Y = 1 and (X'Z')1 = (X'Z') so --> (X'Z') + (XY'Z') The next level is much trickier. It uses the law of A + A'B = A + B (I assume this the same as A' + AB = A' + B ?) again, start by factoring out a common Z' --> Z'(X' + XY') so (X' + XY') = X' + Y' (right?) --> Z'(X' + Y') which is the same as the next to the last line (X'Z') + (Y'Z'), but already reduced. Is that right? I'm a little fuzzy on the second reduction (especially since it skipped straight to the final reduction), but I think that's right.
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## Re: Chap 1 Slide 19: Simplifications of the canonical representation

 found this list of laws: https://www.mi.mun.ca/users/cchaulk/misc/boolean.htmAB' + B = A + B -- redundancy law
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## Re: Chap 1 Slide 19: Simplifications of the canonical representation

 In reply to this post by xedover Hello ! Thank you for the answer. I now understand ! Just a little correction; you wrote : Y' + Y = 0 and (X'Z')0 = (X'Z') so --> (X'Z') + (XY'Z') It should be: Y' + Y = 1                   (Complement law) and (X'Z')1 = (X'Z')     (Identity Law) so --> (X'Z') + (XY'Z') Thanks !
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## Re: Chap 1 Slide 19: Simplifications of the canonical representation

 oh...totally right... I'll make the correction above. thanks.