We were asked to simplify a 2's complement truth table and try to use an xor gate. The term was derived from a Karnaugh map and looked like
A'B + A'C + A'D + AB'C'D'
My next step was pulling out the A'
A'(B+C+D) + A(B'C'D')
Here I run into an issue. I should use an XOR gate, but I'm not sure if I can apply Demorgan's to just the term on the right to do so.
(B'C'D')'' ---> B'+C'+D'
I ended up with this simplified expression
A XOR (B + C + D)
Is that correct? Thanks
First of all, after appealing to DeMorgan's on the right-hand side, you went from $A(B'C'D')$ to $A(B'+C'+D')$, but your application of DeMorgan's rule ends incorrectly.
We can use DeMorgans, on $$A(B'C'D')\equiv A(B+ C+ D)'$$
That gives us $$A'(B+C+D) + A(B+ C+ D)'$$
Do you see how to take it from here?
Finally, yes, you've simplified correctly: $$A \oplus (B+C + D)$$ using the symbol $\oplus$ to designate XOR.