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simplifying simple logic |
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| Feb9-06, 07:16 AM | #1 |
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simplifying simple logic
let's consider this boolean logic experession:
s=x'y'z+x'yz'+xy'z'+xyz can i simplify it to: s=x'y'z+x'yz'+xy'z' as xyz=1 where x,y,z in high logic(1) what's the simplest expression? |
| Feb9-06, 09:37 AM | #2 |
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You have 2 xor's
s=x'y'z+x'yz'+xy'z'+xyz x'y'z + xyz = z(x'y' + xy) = z(x xor y) x'yz' + xy'z' = z'(x'y + xy') = z'(x xor y) add these up and go from there |
| Feb9-06, 12:24 PM | #3 |
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are you sure from that? thanks |
| Feb9-06, 02:28 PM | #4 |
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simplifying simple logic
yea
xy +x'y' is just the complement of (x xor y) (xy' + x'y)' = (x' + y)(x + y') = xy + x'y' |
| Feb9-06, 02:48 PM | #5 |
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s=x'y'z+x'yz'+xy'z'+xyz
x'y'z + xyz = z(x'y' + xy) = z(x xor y)' x'yz' + xy'z' = z'(x'y + xy') = z'(x xor y) so s=(z) xor (y) xor (z) |
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