Fundamentals of Boolean Algebra (6)
Theorem 8 (DeMorgan's Theorem)
(a) (a + b)' = a'b' (b) (ab)' = a' + b'
Generalized DeMorgan's Theorem
(a) (a + b + … z)' = a'b' … z' (b) (ab … z)' = a' + b' + … z'
Examples:
- (a + bc)' = (a + (bc))'
= a'(bc)' [T8(a)]
= a'(b' + c') [T8(b)]
= a'b' + a'c' [P5(b)]
- Note: (a + bc)' ¹ a'b' + c'