y - (a+b+c)(a+b^+c)(a-b^-c^)(a^-b-c^)(a^-b^-c)
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I think that the true answer is C+A′B′C+A′B′and not C+A′BC+A′B as you suggest.
You did well, but could go on with:
C+A′B′C′=(C+A′B′)(C+C′)=C+A′B′C+A′B′C′=(C+A′B′)(C+C′)=C+A′B′
To get hold of the situation you could also make a Venn-diagram on:
(A∩B∩C)∪(A∩B∁∩C)∪(A∁∩B∩C)∪(A∁∩B∁∩C)∪(A∁∩B∁∩C∁)=(A∩B∩C)∪(A∩B∁∩C)∪(A∁∩B∩C)∪(A∁∩B∁∩C)∪(A∁∩B∁∩C∁)=
[[(A∩B)∪(A∩B
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Answer:first simplyfy boolean function then solve it
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