Boolean gate logic

Question

I've got this expression and I believe I have the circuit drawn out correctly however I think the truth table for it is a trick question. There is 4 inputs and 2 of them are A and the other 2 are B and C. Here is what I've drawn out:

Here's what I've done for the truth table.

Answer 1

Your logic expression can be simplified using Boolean Algebra.

Start with the given logic expression:

(ab + ac)'

Factor out a:

(a(b + c))'

Apply De Morgan's law for negation of a conjunction:

a' + (b + c)'

Apply De Morgan's law for negation of a disjunction:

a' + b'c'

Therefore, output is true:

• anytime a is false, OR
• when b is false AND c is false

Here is your truth table with corrections:

Answer 2

A false, output true (regardless of B and C) B and C false, output true (regardless of A) Your truth table is not correct.