If $s_{1}\Longleftrightarrow s_{2}$ then $s_{1}\leftrightarrow s_{2}$ is a tautology? 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.
 A: $\left(s_{1}\Longleftrightarrow s_{2}\right) \leftrightarrow \left(s_{1}\leftrightarrow s_{2}\right)$ and $\left(s_{1}\Longleftrightarrow s_{2}\right) \Longleftrightarrow \left(s_{1}\leftrightarrow s_{2}\right)$ are both tautologies, and both look silly.
A: 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 $v(s_1\leftrightarrow s_2)=1$.
