[Math] About the tautological implication

logic

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?

Best Answer

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).

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