Why is $p \lor ¬q$ true when $p$ is $1$ and $q$ is $1$?

Question

I am studying for the final in my logic class and I am having trouble understanding a problem. The notes say that for the propositional formula: $U = \{¬p ∧ ¬ q, (p ∨ ¬q) ∧ ¬r, p → r\}$ a non-satisfying interpretation would be $p = 1$ $q= 1$ and $r = 0$. I understand why the first and last formulas are false but why is the second one false. I thought that $(p ∨ ¬q)$ would interpret to be $(1 \lor ¬1)$ which one be $(1 \lor 0)$ which would be true since it is saying $1$ or $0$.

Answer 1

In order to satisfy a set of formulas, you need to find an interpretation that makes all formulas true. So, for an interpretation not to satisfy such a set, it suffices for there to be at least one formula that is made false. And that is what is the case here. That is, it need not be the case that all formulas are made false in order for an interpretation to be non-satisfying.

And so you are right: $p \lor \neg q$ is true when both $p$ and $q$ are true. But, since the other two formulas are false, the interpretation is non-satisfying.