How is $[P \text{ AND } (Q \text{ OR } R)] \text{ IFF } [(P \text{ AND } Q) \text{ OR } (P \text{ AND } R)]$ valid?

Question

The image below was taken from some reading materials in a public course on the mathematics of computer science. The text states that the proposition [P AND (Q OR R)] IFF [(P AND Q) OR (P AND R)] is valid.

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

Answer 1

The truth table in your image does not have a column for the value of the entire formula -- only for the subformula on each side of the "IFF".

Since those two columns have the same pattern of T and F, a column for the entire formula would have T all the way.

Answer 2

Compare the last two columns in the table, they are identical, aren't they?

Hence

$$P \land (Q \lor R)$$

and

$$(P \land Q) \lor (P \land R)$$

are equivalent.

Answer 3

The formula in question involves "iff." The statement "A iff B" evaluates to true when A and B are either both true or both false. The truth table shows that the A and B in question are always either both true or both false.

Answer 4

Question: how is this formula 'valid' when clearly it evaluates to False in some of the cases (ex. when P,Q,R are all False)?

$\def\too{\leftrightarrow}$Not just some cases, the clauses evaluate to false in the same cases.   You see that $P\land(Q\lor R)$ is false is exactly when $(P\land Q)\lor (P\land R)$ is false.   Likewise for all true cases.

In all cases $P\land (Q\lor R)$ and $(P\land Q)\lor(P\land R)$ have the same truth value.   They are equivalent.   Then $(P\land (Q\lor R))\too((P\land Q)\lor(P\land R))$ is alwas true because that is what the biconditional means.