X& & ! (x & & y) =
x & & (! x ||! y) =
(x& & ! x)||(x &a...
Computers and Technology, 05.12.2019 09:31 janiyahmcgolley
X& & ! (x & & y) =
x & & (! x ||! y) =
(x& & ! x)||(x & & ! y) =
0 || (x& & ! y) =
x& & ! y
demorgan's
distributive
inverse
identity
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