# Pretty simple NAND gate question

1. Feb 20, 2012

### DrummingAtom

1. The problem statement, all variables and given/known data

I was looking through some examples in my Digital Logic book and I stumbled across one that gave an answer of x1$\uparrow$(x2$\uparrow$x2). The answer I got was x2$\uparrow$(x1$\uparrow$x2).

After making a truth table I'm finding that these are not equal solutions. Unless, I'm doing something wrong.

3. The attempt at a solution

See above. Any help is appreciated.

2. Feb 20, 2012

### Staff: Mentor

Can you write this in words so I can see where NAND comes into it?

3. Feb 20, 2012

### DrummingAtom

The problem asked to turn x1$\overline{x}$2 + x2$\overline{x}$1 into a NAND only circuit. I used decomposition in the last steps to get my answer.

4. Feb 20, 2012

### rcgldr

Can you show the truth table for x1$\overline{x}$2 + x2$\overline{x}$1? I'm not sure the book answer is correct assuming that ↑ means NAND.

5. Feb 20, 2012

### DrummingAtom

I got my question answered today in class. Those two solutions are equivalent. Thanks for the help though.

6. Feb 20, 2012

### rcgldr

I'm somewhat confused by this. You never explained what ↑ means.

Isn't x1$\overline{x}$2 + x2$\overline{x}$1 the same as XOR? Wiki article for XOR implemented with NAND gates:

http://en.wikipedia.org/wiki/NAND_logic#XOR

7. Feb 20, 2012

### DrummingAtom

According to my book, ↑, is a NAND gate operator.

8. Feb 20, 2012

### rcgldr

OK, so again I ask what is the truth table for

x1$\overline{x}$2 + x2$\overline{x}$1

and the truth table for

x1↑(x2↑x2)

or

x2↑(x1↑x2)

9. Feb 20, 2012

### Staff: Mentor

But a poor choice, and bound to lead to confusion.

10. Feb 21, 2012

### Staff: Mentor

Yes, that is exclusive NOR.

x2 NAND x2 is just NOT x2

x2↑(x1↑x2) evaluates to x2 NAND NOT x1

So expressions in post #1 are NOT identical.

Last edited: Feb 21, 2012