Prove that (a+bc) + (a'(b'+c')) = 1

Question

Im trying to prove the boolean expression (a+bc) is the complement of (a'(b'+c')). To do that I need to prove that (a+bc) + (a'(b'+c')) = 1, however I only get as far as:

(a + bc + a')(a + bc + b' + c')
(a + a' + bc)(a + bc + b' + c')
(1 + bc)(a + bc + b' + c')
1(a + bc + b' + c')

What property can I use to reduce the RHS?

Alternatively: $\overline{\bar{a}(\bar{b}+\bar{c})}= \overline{\bar{a}}+\overline{\bar{b}+\bar{c}}=a+bc$ — bigfocalchord, Dec 13 '17 at 00:34

score 0 · Answer 1 · answered Dec 13 '17 at 00:36

0

You're doing very well! Now, recall that $$b'+c'=(bc)',$$ so that we get $$a+bc+(bc)'.$$ Can you take it from there?

answered Dec 13 '17 at 00:36

Cameron Buie

102,994

a + bc + (bc)' = a + 1 = 1; thus (1)(1) = 1 proving the complement? – pstatix Dec 13 '17 at 00:38
Well, not quite. You've done half of the job. You also need to show that $$(a+bc)\bigl(a'(b'+c')\bigr)=0.$$ – Cameron Buie Dec 13 '17 at 00:40
Yea working on that part now, perhaps I meant proving the first part?! – pstatix Dec 13 '17 at 00:42

score 0 · Answer 2 · answered Dec 14 '17 at 22:55

$$a \lor bc \lor \overline a(\overline b \lor \overline c)=1$$ I use this rule: $a=a \lor a \overline b=a \lor a \overline c$. $$[a \lor bc \lor \overline a \overline b \lor \overline a \cdot \overline c] \lor a \overline b \lor a \overline c =1 $$ You should remember that $ x \lor \overline x = 1$. $$ a \lor \overline b \lor \overline c \lor bc =1$$ $$ a \lor \overline b \lor \overline c \lor bc \lor \overline b c =1$$ $$ a \lor \overline b \lor \overline c \lor c =1$$ $$ a \lor \overline b \lor 1 =1$$ So, this function is always true.

Prove that (a+bc) + (a'(b'+c')) = 1

2 Answers2