Re: Boolean algebra
The conclusion seems to be correct, as evident from the truth tables:
Original expression, first term makes it false for 11, second for 10 and third for 01.
For 00, the original gives true.
so, reduction to a'b' is justified
(a' + b')(a' + b)(a + b') =
(a' + b'a' + a'b + b'b)(a + b') =
(a' + b'a' + a'b + 0) =
(a' + b'a' + a'b)(a + b') =
a'a + b'a'a + a'ba + a'b' + b'a'b' + a'bb' =
0 + 0 + 0 + a'b' + b'a' + 0 = a'b' + a'b' =
a'b'
--
The rich, as usual, are employing the elected.
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