Simplification of boolean expression: $(A+BC’) (A(A+B’C)) + A’ + B’$

Question

Good evening,

As the title suggests, I need help to minimize a boolean expression and verify it with the $K$ map. I tried to solve it and the result is equal to $1$ but I can't get the same result with the $k$ map. I therefore believe that i made some mistakes during minimization. Could you please help?

The boolean expression is : $(A+BC') (A(A+B'C)) + A' + B'$

Thanks to advance.

Answer 1

A K-map is not really needed here. Let's take a look at just part of the first term in your expression:

$$A(A+B'C)$$

Here, we can distribute the 'and', producing $A + AB'C$. We then have the 'or' of $A$ and something that depends on $A$, so the result is simply $A$. Let's substitute back into the larger expression:

$$(A+BC')A + A' + B'$$

We perform the same simplification on the first term.

$$A + A' + B' = 1 + B' = 1$$

If you produce a K-map of the function, you'll find that it will be eight cells of 1's.