First, of course, mathematicians and logicians often do use truth tables, so it is incorrect to suggest that they are not used. (I won't comment on your reading of Principia Mathematica, beyond saying that that text is not meant to be taken as a piece of pedagogy.)
But second, the truth table method is not a feasible method for large propositions. This is because for a formula with $n$ free variables, the truth table is an object with $2^n$ rows, exponentially large in size. But many logical expressions nevertheless have comparatively short derivations showing them to be tautological.
For example, $$(p_0\vee \neg p_0)\wedge(p_1\vee \neg p_1)\wedge\cdots\wedge(p_{100}\vee\neg p_{100})$$ has a truth table with $2^{101}$ rows, but there is a simple, obvious derivation that allows us to see it as taugological without calculating this table.
Thus, although when there are just two propositional variables, I don't mind computing a truth table, or even for three, it is often easier with more to use more focused reasoning.
In the context of propositional logic, a model is simply an assignment of truth values to the basic propositional variables. The statement $p\land q$ is true in any model that satisfies both $p$ and $q$, i.e., in which $p$ and $q$ are both true; the statement $p\lor q$, on the other hand, is true in any model that satisfies at least one of $p$ and $q$. This information is precisely what you’re collecting when you write out a truth table.
To see whether a formula $F_2$ is a logical consequence of a formula $F_1$, write out their joint truth table. Find the rows in which $F_1$ is true; those are essentially the models that satisfy $F_1$. Check to see whether $F_2$ is also true in all of those rows/models; if it is, then every model that satisfies $F_1$ automatically satisfies $F_2$, and $F_2$ is therefore a logical consequence of $F_1$. Here’s an example.
$$\begin{align*}
&F_1:\quad\big(p\to(q\lor r)\big)\land\lnot r\\
&F_2:\quad p\to q
\end{align*}$$
Here’s the combined truth table:
$$\begin{array}{c|c}
p&q&r&q\lor r&p\to(q\lor r)&\lnot r&p\to(q\lor r)\big)\land\lnot r&p\to q\\ \hline
T&T&T&T&T&F&F&\color{blue}{T}\\
\color{green}{T}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\
T&F&T&T&T&F&F&F\\
T&F&F&F&F&T&F&F\\
F&T&T&T&T&F&F&\color{blue}{T}\\
\color{green}{F}&\color{green}{T}&\color{green}{F}&T&T&T&\color{red}{T}&\color{blue}{T}\\
F&F&T&T&T&F&F&\color{blue}{T}\\
\color{green}{F}&\color{green}{F}&\color{green}{F}&F&T&T&\color{red}{T}&\color{blue}{T}
\end{array}$$
The $T$’s for $F_1$ are in red, and those for $F_2$ are in blue. As you can see, in every row in which $F_1$ is true, $F_2$ is also true; the truth values of $p,q$, and $r$ in those rows are in green. $F_1$ holds in precisely those models in which $p$ and $q$ are true and $r$ is false, or $q$ is true and $p$ and $r$ are false, or all three are false. And in all such models $F_2$ holds as well, so $F_2$ is a logical consequence of $F_1$.
Note, though, that $F_2$ holds in some models in which $F_1$ does not hold, e.g., those in which $p,q$, and $r$ are all true. Thus, $F_1$ is not a logical consequence of $F_2$, and therefore $F_1$ and $F_2$ are not logically equivalent.
Best Answer
You are missing a quantifier. You are comparing a statement of the form "Everyday X holds" to "Everyday not X holds". There is no reason those should be opposites. It is possible for those both to be false, just allow for some days X and some days not X.
The proper comparison would be
vs
Work out what goes into ???? and I think you'll have it.