What is the mathematical difference of the congruent and iff operators on two boolean values? Denoting boolean values (0 or 1) as $\mathbb{B}$, if $a, b \in \mathbb{B}$, is $a \equiv b$ mathematically identical to $a \Longleftrightarrow b$, given that they have identical truth tables?
 A: There is another way of thinking about the logical operator $\iff$. If we think of the boolean value false as $0$ and the boolean value true as $1$, then the value of an equation involving boolean variables $a$ and $b$ like this one $a\iff b$ can be though of as $a+b+1(mod 2)$. 
Let us check that indeed these two equation compute the same thing. 
\begin{array} {|r|r|}\hline a & b & a+b+1(mod 2) & a \iff b \\ \hline 1 & 0 & 0 & 0 \\ \hline 1 & 1 & 1 & 1 \\ \hline 0 & 0 & 1 & 1 \\ \hline 0 & 1 & 0 & 0 \\ \hline  \end{array}
This way of viewing the boolean operator $\iff$ is very useful sometimes. For example, consider the claim that a boolean expression involving a finite number of variables and that each variable is used an even number of times, and the logical operator is restricted also to only the $\iff$ operator, then the boolean expression is a tautology, namely, the boolean expression is always true no matter the value $\{0,1\}$ that you assign to the variables. 
If you were to use the definition of the logical operator $\iff$, it is probably not going to be very easy to prove the claim. But if you view the logical expression $a \iff b$ as $a+b+1(mod2)$, then the problem could be much easier. Namely, if you first change your boolean expression in the modular addition format and get rid of the parenthesis, you will get the sum of each variable an even number of times. The sum of the variable will therefore be equals to $0(mod2)$. Futhermore , each variable appear an even number of times means that there are an odd number of $\iff$ operators in the original boolean expression. When you converted the boolean expression into the expression involving sums of the variable + $1$, you essentially converted the odd number of $\iff$ operators into the sum of odd numbers of $1$'s plus the variables. We know that the sum of the variables is $0(mod2)$ and the sum of odd number of $1$'s is $1(mod2)$, therefore, the value of your converted expression is also $1(mod2)$. And since $1$ correspond to the boolean value True, you know that your expression is always true, hence a tautology. 
Therefore, you see that this is a powerful technique of converting a boolean operator into  arithmetic in mod2.
