If $s_{1}\Longleftrightarrow s_{2}$ then $s_{1}\leftrightarrow s_{2}$ is a tautology?

Question

I see that it's not always the case for $s_{1}\leftrightarrow s_{2}$ is a tautology. As if I have $s_{1}:p$ and $s_{2}:q$, I have the following truth table: $$ \begin{array}{c|c|c|c|c|c|c} p & q & \neg p & \neg q & (\neg p \lor q) & (\neg q \lor p) & (\neg p \lor q)\land(\neg q \lor p) \\ \hline 0 & 0 & 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 1 & 0 & 1 & 0 \\ 1 & 1 & 0 & 0 & 1 & 1 & 1 \\ \end{array} $$

(In the truth tables I replace $p\leftrightarrow q$ with $(\neg p \lor q)\land(\neg q \lor p)$, as $p\leftrightarrow q\Longleftrightarrow(\neg p \lor q)\land(\neg q \lor p)).$

But there are instances for which they are, i.e., let $s_{1}: p \lor \neg p$ and $s_{2}: q \lor \neg q$, we have the following truth table

$$ \begin{array}{c|c|c|c|c|c|c|c|c} p & \neg p & q & \neg q & p\lor \neg p & q \lor \neg q & (\neg p \lor q) & (\neg q \lor p) & (\neg p \lor q) \land (\neg q \lor p) \\ \hline 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 0 & 1 & 1 & 1 & 1 & 1 \\ \end{array} $$

So from this result can I conclude the following two statements:

$\textbf{(1)}$ If $s_{1},s_{2}$ are statements and $s_{1}\leftrightarrow s_{2}$ is a tautology, then $s_{1} \Longleftrightarrow s_{2}$?

$\textbf{(2)}$ If $s_{1}\Longleftrightarrow s_{2}$ then $s_{1}\leftrightarrow s_{2}$ is a tautology?

Thanks.

Answer 1

$\left(s_{1}\Longleftrightarrow s_{2}\right) \leftrightarrow \left(s_{1}\leftrightarrow s_{2}\right)$ and $\left(s_{1}\Longleftrightarrow s_{2}\right) \Longleftrightarrow \left(s_{1}\leftrightarrow s_{2}\right)$ are both tautologies, and both look silly.

Answer 2

You can indeed infer $\textbf{(1)}$ and $\textbf{(2)}$ from looking at the truth tables, but to write it requires some care. Also it's not good practice to say $s_1=p$, because $p$ is usually reserved for propositional symbols and $s_1$ can be a complex formula, but this isn't really relevant. Below is how I would have done it.

By definition, $s_1\iff s_2$ is an abbreviation of $s_1\implies s_2$ and $s_2\implies s_1$, or in an alternative notation, $s_1\models s_2$ and $s_2\models s_1$. I will be using the latter.

For $\textbf{(1)}$, assume $s_1\leftrightarrow s_2$ is a tautology. This means, by definition of tautology, that for every valuation $w$ one has $w(s_1\leftrightarrow s_2)=1$. Since by definition of valuation one has $$w(s_1\leftrightarrow s_2)=\min\left(\max(1-w(s_1), w(s_2)), \max(w(s_1), 1-w(s_2))\right),$$ it follows that $\max(1-w(s_1), w(s_2))=1=\max(w(s_1), 1-w(s_2))$.

Now to prove that $s_1\models s_2$ take any valuation $v$ and suppose that $v(s_1)=1$. (The goal is to prove that $v(s_2)=1$). Since $\max(1-v(s_1), v(s_2))=1$ and $1-v(s_1)=0$, one gets $v(s_2)=1$, as desired.

The proof of $s_2\models s_1$ is similar.

For $\textbf{(2)}$, suppose $s_1\models s_2$ and $s_2\models s_1$. (The goal is to prove that for every valuation $v$, $v(s_1\leftrightarrow s_2)=1$). Let $v$ by a valuation. One has $$v(s_1\leftrightarrow s_2)=\min\left(\max(1-v(s_1), v(s_2)), \max(v(s_1), 1-v(s_2))\right).$$

Now either $v(s_1)=1$ or $v(s_1)=0$. In the first case, since $s_1\models s_2$, $v(s_2)=1$ follows. Consequently $v(s_1\leftrightarrow s_2)=1$. In the case $v(s_1)=0$, because $s_2\models s_1$, it can't be the case that $v(s_2)=1$, thus $v(s_2)=0$ and $v(s_1\leftrightarrow s_2)=1$.