# Truth table for X = A.(B+C)'?

I have been beating my head on this for hours. I'm pretty certain that I've done it correctly, but my Quartus II simulation seems to disagree.

My Boolean expression: X = A.(B+C)'

My truth table (supposedly incorrect):

A   B   C   X
1   0   0   1   (0 is supposedly the correct output)
0   0   0   0
1   1   0   0
0   1   0   0
1   1   1   0   (1 is supposedly the correct output)
0   1   1   0
1   0   1   0
0   0   1   0


The thing is, I if my equation is right then I don't see how they get those values. So here's my Quartus II block diagram...

Well, I can't actually post the screen shot of the circuit, but it was an output of X with a wire to a AND2 gate which has wires to a NOR2 gate and Input A. The NOR2 gate has wires to inputs B and C.

I'm about to go crazy here. Can anyone provide some insight into where I'm erring? Perhaps my Boolean expression is incorrect for the diagram? Or perhaps I'm incorrectly calculating the Boolean expression?

Any help is much appreciated.

-
Your table is perfect. You need $A = 1$ and $(B+C) = 0$ for $X = 1$. –  hjpotter92 Jan 27 '13 at 6:26
I am a little confused with your notation. Does $'$ represent not? What is the order of precedence as written? Regards –  Amzoti Jan 27 '13 at 6:28
@Amzoti: Judging by the circuit described in the question, it’s $A\land\neg(B\lor C)$, and Abe’s truth table is indeed correct. The answer that’s supposed to be correct goes with $A\land B\land C$, which clearly doesn’t match the circuit. My tentative guess is that whoever prepared the answer key applied De Morgan to get $A\cdot(B'\cdot C')$ and inadvertently dropped the primes. –  Brian M. Scott Jan 27 '13 at 6:29
@BrianM.Scott: I really admire your problem solving skills if you don't mind me saying. So, wouldn't $1 and (0 + 0)' = 1 and 1= 1$? Regards –  Amzoti Jan 27 '13 at 6:34
@Amzoti: Thank you. And yes, $1\land(0+0)'=1\land 1=1$, just as Abe has it in the truth table. –  Brian M. Scott Jan 27 '13 at 6:36