Page 32 - Computer science 868 Class 12
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Example 3: Minimise the following expression using laws of Boolean algebra.
Q.(Q'+P).R.(Q+R) [ISC 2012]
Solution:
= (Q.Q'+Q.P).R [Distributive and absorption laws]
= (0+Q.P).R [Complement law]
= (Q.P).R [Properties of 0 and 1]
= P.Q.R [Distributive law]
Example 4: Prove De Morgan’s law algebraically.
Solution:
De Morgan’s Law
(i) (A+B)' = A'.B'
(ii) (A.B)' = A'+B'
To prove (A+B)'=A'.B', we have to apply Complement law X.X'= 0.
According to De Morgan’s Law, A'.B' is the complement of (A+B).
Then according to Complement law, (A+B).A'.B' = 0 should be satisfied.
So, LHS = (A+B).A'.B'
= A.A'.B' + B.A'.B' [Distributive law]
= 0+0 [Complement law]
= 0
To prove (A.B)' = A'+B', we use the other Complement law which states that X+X'=1.
According to De Morgan’s Law, A'+B' is the complement of (A.B).
Then according to Complement law, (A.B) + A'+B' = 1 should be satisfied.
So, LHS = (A.B) + A'+B'
= (A.B) + (A'+B')
= (A'+B'+A).(A'+B'+B) [Distributive law]
= (1+B').(A'+1) [Complement law]
= 1.1
= 1
Thus (A+B)' = A'.B' and (A.B)' = A'+B' proved.
Example 5: Minimise the following expression using laws of Boolean algebra.
(P+Q).(P'+Q).(P'+Q')
Solution:
= (P+Q).(P'+Q).(P'+Q')
= (Q + P.P').(P'+Q') [Distributive law]
= Q.(P'+Q') [Complement law]
= P'.Q + Q.Q' [Distributive law]
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