Question

simplify A[B+C'(AB+AC') ']

Answer 1

Answer:

Since we know that (A+B)' = A'B', we can write C'(AB+AC')' as (C+AB+AC')'. So the expression reduces to A(B+(C+AB+AC')').

Again using (A+B)' = A'B', we can rewrite B+(C+AB+AC')' as (B'(C+AB+AC'))'. This when simplified by using distributive property leads to (B'C+B'AB+B'AC')'. But B'AB is always FALSE or 0. So this is effectively (B'C+B'AC')'.

So now the expression is A(B'C+B'AC')'. Again using (AB)' = A'+B', we can rewrite A(B'C+B'AC')' as (A'+B'C+B'AC')'. Now, write B'C as B'C(A+A'), as 1 can be written AS A+A', we can rewrite it as (A'+B'CA+B'CA'+B'AC')'.

Here B'CA+B'AC' is B'A(C+C') which is same as B'A. Hence the expression reduces to (A'+B'A+B'CA')'. Now here A'+B'CA' can be written as A'(1+B'C). Since 1+anything is 1 itself, this reduces to A.

So we now have our expression as (A'+B'A)'. Now we know that A+A'B = A+B. So our expression reduces to (A'+B')' and this finally reduces to AB.

So the simplified form of the expression is AB.

Note: It is important to remember that A+A’B = A+B as it is quite useful. How do we get this? Write LHS as (A'(A'B)')'. This can be written as (A'(A+B'))'. This can be written as (A'A+A'B')'. But A'A is 0. So the expression simplifies to (A'B')'. This finally reduces to A+B.

Thank you.

Explanation:

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