About the tautological implication

Question

Definition: Let $p$ and $q$ be two compound statements. I read that, If $p\implies q$ is a tautology, then $q$ is said to be a logical consequence of $p$.

Furthermore, it notes that the statement $p\implies q$ is automatically true when $p$ is false, and saying that $p\implies q$ is a tautology actually means that $q$ is true, when $p$ is true.

I do not really understand the last sentence (the bold ones). Can $p\implies q$ not be a tautology, when $p$ is false, or when $p$ and $q$ are both false?

For example, let $p$ and $q$ be simple statements, and we know that the compound statement $[(p\implies q)\wedge p]\implies q$ is a tautology, which can be shown by making a truth table, see for example URL (generate "((p>q)&p)>q" without the "-signs).

Does the note say that we only need to check when $(p\implies q)\wedge p$ is true, which only happens when $p$ and $q$ are true (as shown in the truth table), and we forget the rest of the combinations?

Answer 1

Furthermore, it notes that the statement $\def\implies{\to}p\implies q$ is automatically true when $p$ is false, and saying that $p\implies q$ is a tautology actually means that $q$ is true, when $p$ is true.

I do not really understand the last sentence (the bold ones). Can $p\implies q$ not be a tautology, when $p$ is false, or when $p$ and $q$ are both false?

As stated, $p\to q$ is true when $p$ is false; that is a sure thing.   Thus we only need to examine what happens to $p\to q$, when $p$ is true.   Hence $p\to q$ will be a tautology if we cannot have $q$ be false while $p$ is true.

For example: $p\to (r\to p)$ is a tautology because $r\to p$ is alway true when $p$ is true.   Now $r\to p$ may be false when $p$ is false (ie when $r$ is true), but that doesn't matter because $p\to (r\to p)$ is true when $p$ is false whatever $r\to p$ might be.

On the other hand $p\to\lnot p$ is not a tautology because, clearly, $\lnot p$ may$^\star$ be false when $p$ is true. ($^\star$ in fact, it must be so).

Answer 2

$p \to q$ is TRUE either when $q$ is TRUE or when $p$ is FALSE.

Equivalently, $p \to q$ is FALSE exactly when $p$ is TRUE and $q$ is FALSE.

A formula is a tautology when it is always TRUE.

Thus, in order for $A \to B$ to be a tautology, we have to exclude the case : $A$ TRUE and $B$ FALSE.

In order to verify that a fromula is a tautology we have to build a truth table and check that the formula is TRUE in every row.

In order to verify that $B$ is a tautological consequence of $A$ we have to build a truth table for both formulas $A$ and $B$ and check that in every row where $A$ is TRUE also $B$ is TRUE.

Answer 3

Does the note say that we only need to check when $(p\implies q)\wedge p$ is true, which only happens when $p$ and $q$ are true (as shown in the truth table), and we forget the rest of the combinations?

No. Using only 6 basic, self-evident principles of natural deduction (Intro $\land$, Elim $\land $, Intro $\to$, Elim $\to$, Intro $\neg$, Elim $\neg\neg$), we can actually prove each of the following (each corresponding to a line of the truth table):

Line 1: $A \land B \to (A \to B)$

Line 2: $A\land \neg B \to \neg(A \to B)$

Line 3: $\neg A \land B\to (A\to B)$

Line 4: $\neg A \land \neg B \to (A \to B)$

Thus we can derive the entire truth table for material implication.