Question

in solving bollean algebra problem
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x=a.b+c+a.c+b
here there is a whole bar on a.b+c
and also a bar on c.there is another bar +a.c where c have also bar. plzzz help me

Answer 1

o.O

Answer 2

Lahori, right?

Answer 3

\[\bar a + a = 1\]

Is that right?

Answer 4

and (c+a)(c+b)= c + ab

Is that the type of logic we are talking about?

Answer 5

\[\overline{a.b+c}+\overline{c}+\overline{a.\bar c}+b\] ?

Answer 6

without the c in the middle I think

Answer 7

\[\overline{a.b+c}+\overline{a.\bar c}+b\]

Answer 8

This expression:
not(a and b or not(c)) or b or a and not(c)

Is equivalent to (see wolfram below):
a or b or c

Which needs to be shown with boolean algebra (see a later post made, which seems to refer to this question).


http://www.wolframalpha.com/input/?i=not%28a+and+b+or+not%28c%29%29+or+b+or+a+and+not%28c%29

Answer 9

I hate those equation because I'm ALWAYS making careless mistakes
\[\overline{a.b+\bar c}+\overline{a.\bar c}+b = \overline{a.b}.c+\bar a+c+b=c.(\overline{a.b}+1)+\bar a + b=c.1 +\bar a +b=\bar a+b+c\]

Answer 10

Is it \[\overline{a.b+\bar c}+\overline{a.\bar c}+b\] or \[\overline{a.b+\bar c}+a.\bar c+b\] ?

Answer 11

I think it's the 2nd one you posted, and that you are supposed to show it is equivalent to A+B+C (see the next question that Muhammad posted).

According to wolfram, the 2nd expression you posted IS equivalent to A+B+C .

Answer 12

Yep, the second IS A + B + C

Here is the simplification (at least my try)
\[\overline{a.b+\bar c}+a.\bar c+b = \overline{a.b}.c+a.\bar c+b=(\bar a +\bar b).c+a.\bar c+b=\bar a.c+\bar b.c+a.\bar c +b\]\[=(\bar a.c+a.\bar c) +(\bar b.c+b)=a \oplus c +b +c = (a \oplus c +c) +b= a + c +b\]