Next, let's address the third entry in the truth table.  In that situation, X = 1, and Y = 0.

• If Y = 0, then Q = 1.  We know that if either input to a NAND gate is 0, the output is 1.  That's the same as before.  However, this time, we take advantage of knowing that Y = 0.  Before, it was X that was equal to zero, but X = 1 here and it doesn't help us get started.

• Now. try to take advantage of the knowledge that Q = 1.  If Q = 1 AND X = 1, then P = 0.

• That gives us the third entry in the truth table above.  Here's the truth table with what we have figured out so far.

X	Y	P	Q
0	0	1	1
0	1	1	0
1	0	0	1
1	1

        Finally, we have the case where X = 1 and Y = 1.

• If X = 1, then we need to know Q to determine P.

• If Y = 1, then we need to know P to determine Q.

• We are stuck!  There is no obvious way to proceed!

        There is one way to proceed.  We could just assume that P = 1, for example.  Let's try that.

• If Y = 1, and P = 1, then we know that Q = 0.

• If we know that Q = 0, it is an input  to the top NAND gate, so the output there is 1.  Thus, P = 1.  That's what we assumed to start with, so we don't get a contradiction.

• All is copasetic!  Or is it?

• Here is the truth table.

 

X	Y	P	Q
0	0	1	1
0	1	1	0
1	0	0	1
1	1	1	0