Sum of products expansion of basic Boolean function: $ F(x,y) = \bar{y} $

Question

So I have a question about this very basic-looking sum of products expansion. My professor has this particular example in his lecture slides but I can't quite wrap my head around this.

I don't understand how the 'bar' or negation can be removed from $\bar{y}$ during the application of the Unit Property from step 2 to step 3 in the Boolean function below. This also raises the question: Is $ \bar{y}\cdot1$ also logically equivalent to what I originally thought: $\bar{y}\cdot(x+\bar{x}) $ ?

1. $ F(x,y) = \bar{y} $
2. $ F(x,y) = \bar{y} \cdot 1$
3. $ F(x,y) = y \cdot (x+\bar{x})$
4. $ F(x,y) = xy + \bar{x}y$

Answer 1

Thank you Senex, it turns out that it was a misprint in his slides, I sent him an email and he quickly realized the truth table didn't match up.

So the correct answer is $ F(x,y) = x\cdot \bar{y} + \bar{x}\cdot \bar{y} $ .