$ab+(ac)'+ab'c(ab+c) = 1$? how ?
Asked
Active
Viewed 34 times
-1
-
$a$, $b$, $c$ are either $0$ or $1$ and there are only 8 combinations to check...... – Mar 21 '16 at 06:25
-
How? Use a truth table, and you'll see. – barak manos Mar 21 '16 at 08:57
1 Answers
2
$$a b+ (ac)'+a b' c (a b+c)=$$ $$=a b +(ac)'+ a^2 b b' c+a b' c^2=$$ $$=a b+(ac)'+a^2\cdot 0\cdot c +a b' c^2=$$ $$=a b+(a c)'+a b' c^2=$$ $$=a b+(a c)'+a b' c=$$ $$= a b \cdot 1+(a c)'+a b' c=$$ $$= a b(c+c')+(ac)' +a b' c=$$ $$=a b c+a b c' +(ac)'+a b' c=$$ $$=a (b+b')c+a b c'+(a c)'=$$ $$=a\cdot 1 \cdot c+a b c'+(a c)'=$$ $$=a c +a b c'+(a c)'=$$ $$=[a c+(ac )']+a b c'=$$ $$=1+a b c'=1.$$
DanielWainfleet
- 57,985