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

Answers: 3

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X& & ! (x & & y) =
x & & (! x ||! y) =
(x& & ! x)||(x &a...

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X& & ! (x & & y) = x & & (! x ||! y) = (x& & ! x)||(x & & ! y) = 0 || (x& & ! y) = x& & ! y demorgan's distributive inverse identity

Answers

Another question on Computers and Technology

X& & ! (x & & y) =
x & & (! x ||! y) =
(x& & ! x)||(x & & ! y) =
0 || (x& & ! y) =
x& & ! y
demorgan's
distributive
inverse
identity