Question

in a boolean algebra prove that (i)(a+b)'+(a+b')=a' (ii)(a+b).(a'+c)=a'.b+a.c.

*Answer *​

Answer 1

Answer:

ii.)

is correct option

please check

Answer 2

Theorem 2. (A+B)’ = A’B’

The second theorem states that the compliment of a sum is equal to the product of the

compliments.

This theorem also show that NOR gate and the bubbled AND gate are equivalent.

The compliment of a logical expression can be obtained by the following steps using De

Morgan’s theorem:

Step1: First of all remove the overall NOT sign

Step 2: Replace all ANDs to ORs and all ORs to ANDs

Step 3: Compliment all the individual variables given in the logical expression.

These two theorems can be proved taking A=0,1 and also B=0,1

When A=0 and B=0, then (AB)’ = (0.0)’ = 0’=1 ; A’=1 and B’=1, then A’+B’=1+1=1:

(A+B)’ = (0+0)’ = 0’ =1; A’B’=1.1=1

When A=0 and B=1, then (AB)’ = (0.1)’ = 0’ =1; A’=1 and B’=0, then A’+B’=1+0=1:

(A+B)’ = (0+1)’=1’=0; A’B’=1.0=0

When A=1 and B=0, then (AB)’=(1.0)’=0’=1; A’=0 and B’=1, then A’+B’=0+1=1:

(A+B)’=(1+0)’=1’=0; A’B’=0.1=0

When A=1 and B=1, then (AB)’=(1.1)’=1’=0; A’=0 and B’=0, then A’+B’=0=0=0:

(A+B)’=(1+1)’=1’=0; A’B’=0.0=0

Thus all the above examples give an authentic proof of the two theorems.