By using the Boolean algebra simplify the following expressions

Question

I. F(A,B,C,D) = A + A'B + A'B'C + A'B'C'D + A’B’C’D’
              = A + A'B + A'B'C(D + D') + A'B'C'D + A'B'C'D'      inverse law
              = A + A'B + D(A'B'C) + D'(A'B'C) + A'B'C'D + A'B'C'D'   distributive law
              = A + A'B + C(A'B'D) + C(A'B'D') + C'(A'B'D) + C'(A'B'D')   associative law
              = A + A'B + C(A'B'D) + C'(A'B'D) + C(A'B'D') + C'(A'B'D')
              = A + A'B + (C + C')(A'B'D) + (C + C')(A'B'D')      distributive law
              = A + A'B + A'B'D + A'B'D'              inverse law
              = A + A'B + D(A'B') + D'(A'B')          associative law
              = A + A'B + (D + D')(A'B')              distributive law
              = A + A'B + A'B'                        inverse law
              = A + B(A') + B'(A')                    associative law
              = A + (B+B')(A')                        distributive law
              = A + A'                                inverse law
              = True                                  inverse law

Not entirely sure if this is correct. Can someone check it and confirm for me? Thanks!

Answer 1

Everything is correct, good job!

But the beginning can be done a little faster:

$$A + A'B + A'B'C + A'B'C'D + A'B'C'D'=$$

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

$$A + A'B + A'B'C + A'B'C'=$$

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

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

$$A + A'(B + B')=$$

$$A + A'=$$

$$True$$

... which pretty much follows the same method you are using, just a little faster.

But here is an alternative method, based on:

Reduction

$P + P'Q=P+Q$

Applying Reduction on your statement, we get:

$$A + A'B + A'B'C + A'B'C'D + A'B'C'D'=$$

$$A + B + B'C + B'C'D + B'C'D'=$$

$$A + B + C + C'D +C'D'=$$

$$A + B + C + D +D'=$$

$$A + B + C + True=$$

$$ True$$

Make sure to put Reduction in your 'Boolean Algebra Toolbox'!