Why is the distributive law incorrect here?

Question

I am trying to simplify:

$(p \lor \neg q) \land (p \lor q) $

One thing, I identify from the table is this:

$(p \lor q) \land (p \lor r) $ is second distributive law which becomes $p \lor (q \land r) $

My steps:

1. $(p \lor \neg q) \land (p \lor q) $
2. $p \lor (q \lor \neg q) $
3. $p$

I watched this video: https://www.youtube.com/watch?v=fXfv1Ru5tzE&pp=ygUebG9naWMgc2ltcGxpZmljYXRpb24gdGF1dG9sb2d5

I was told that, my approach is wrong above.

Answer 1

My steps:

1. $(p \lor \neg q) \land (p \lor q) $
2. $p \lor (q \lor \neg q) $

No. That should be $p \lor (q \color{red}\land \neg q) $

3. $p$

No. First of all, your $q \lor \neg q$ is equivalent to $\top$, and so you would get $p \lor \top$. Second, $p \lor \top$ is equivalent to $\top$, rather than $p$. And third: make sure to show that step.

So, correcting this:

1. $(p \lor \neg q) \land (p \lor q) $
2. $p \lor (q \land \neg q) $
3. $p \lor \bot $
4. $p$

Answer 2

Solution:

1. $(p \lor \neg q) \land (p \lor q)$
2. $p \lor (q \land \neg q) \quad \textit{Second Distributive Law}$
3. $p \lor F \quad \textit{First Inverse Law}$
4. $p$

First Inverse Law: $p \land \neg p \equiv F$