Could somebody help me make sense of how to use the Boolean rules to simplify this expression? $$(x'+(yz)')(x + z')'$$ I used distributivity to get $$(x'+y')(x'+z')(x'+z)$$ I don't know if that was the right path to go down, or where to go from here. Thanks.
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What did you try? – Parcly Taxel May 25 '17 at 03:46
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I used distributivity to get (x' + y')(x' + z')(x' + z). I don't know if that was the right path to go down, or where to go from here. – user433367 May 25 '17 at 03:48
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my apologies, not awfully privy to what needs to shown when asking a question, i hope thats been alleviated – user433367 May 25 '17 at 03:51
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I'd rather take this from the start: $$(x'+(yz)')(x+z')'$$ Apply De Morgan's laws to $(yz)'$ and $(x+z')'$: $$=(x'+y'+z')zx'$$ Distribute: $$=zx'x'+zx'y'+zxz'$$ $x'x'=x'$ and $zz'=0$: $$=zx'+zx'y'$$ $zx'$ absorbs $zx'y'$: $$=zx'$$ This is the simplest form.
Parcly Taxel
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