Show that $p \to q$ is a tautology if and only if $P \subseteq Q$.

Question

I really don't know how to approach this.

I wrote out the truth table for $p' + q$ in which each row has a value of $1$ apart from when $p = 1$ and $q = 0$.

Intuitively if $q = 0$ then $Q$ is a null set. And if $P\subseteq Q$ then that means $\{1\}\subseteq \{0\}$ which is false.. so this row doesn't apply.

All other rows have a value of $1$, and so the said proposition is a tautology.

Please point out if/where I'm wrong in this and how to approach this more mathematically.

Thanks.

Answer 1

$P$ and $Q$ ate the truth sets for $p$ and $q$ respectively, i.e. the sets of valuations that make $p$ and $q$ True respectively.

To say that $p → q$ is a tautology means that there is no valuation such that $p$ is True and $q$ is False, i.e. in every valuation where $p$ is True also $q$ is True.

Thus, the truth set $P$ is a subset of $Q$.

The same for the "only if" part.