Let's suppose we define a statement form (statement hereafter) as follows:
1) All lower case letters of the Latin alphabet are statements.
2) If $\alpha$ is a statement, then $\alpha$$\lnot$ is statement.
3) If $\alpha$ and $\beta$ are statements, then $\alpha$$\beta$$\land$, $\alpha$$\beta$$\lor$, $\alpha$$\beta$→, and $\alpha$$\beta$≡ (this definition has elegance when working with truth tables).
Though I don't know this text, I would guess that the author would say that all sub-statements or proper sub-statements qualify as component statements of a statement. In other words, if we have a formula $\alpha$ a "component statement" is a statement which appears within $\alpha$.
Now suppose we have a statement which is not a variable or constant, like ab≡c$\lor$a$\land$. I will hope that you find it clear that ≡ does not qualify as a component statement of ab≡c$\lor$a$\land$, nor does b≡. The component statements of ab≡c$\lor$a$\land$ are "a", b, ab≡, c, ab≡c$\lor$, and ab≡c$\lor$a$\land$ (if the statement itself gets allowed to qualify as a component of itself, the last string here listed will qualify as a component statement, if not, then the last statement does not qualify as a component statement.) Consequently, a component statement variable, I would think, comes as nothing more than a component statement which also qualifies as a variable.
When do component statement variables come as identical? When they have the same form. Thus, this passage "Two statements are called logically equivalent if, and only if, they have logically equivalent forms when identical component statement variables are used to replace identical component statements." probably implies that
ab≡c$\lor$a$\land$ is logically equivalent to xb≡c$\lor$x$\land$, as well as fb≡c$\lor$f$\land$, but NOT yb≡c$\lor$z$\land$, because "y" and "z" do not have the same form, but "a" and "a" do have the same form. That may seem trivial, but I will point out that when deriving theses (theorems of the object language) substitutions for variables has to occur uniformly. In other words, substitution has to occur for all instances of the variable in a statement.
If you were to put that example into words, you probably would say something like "the first is equivalent to the second or the third and the first." (it isn't clear in words what this means exactly because we don't have a way to associate the words representing the connectives here naturally). But, you would not say "the second is equivalent to the first or the third and the second" because by doing such you immediately have a second variable appearing before the first variable in that statement. Nor would you say "the first is equivalent to the second or the third and the second," and mean what you said with the first example of this paragraph, because if you did say "the first is equivalent to the second or the third and the second," and they were to mean the same proposition, the first would become the second and it becomes permissible to say "the second is..." which leads back to the other absurdity I spoke about in this paragraph.
The first definition implies the second, but the second does not imply the first. Consequently, he can say that the second definition comes as a special case of the first, and he hasn't contradicted himself.
If you have trouble following the definition given in the first paragraph, you might want to read this Wikipedia.
By proposition 1.1.a, $B⇒C$ and $C⇒B$ are tautologies.
No, they are not tautologies. Tautologies are statements that are true for every truth value of the atoms.
You then use the definitions given to complete the truth table. Here's mine.
$$\begin{array}{|c|c|} \hline B & C & B\to C & C\to B & B\to C \wedge C\to B & B\leftrightarrow C
\\[1ex] \hline \top & \top & \top & \top & \top & \top
\\[0ex] \top & \bot & \bot & \top & \bot & \bot
\\[0ex] \bot & \top & \top & \bot & \bot & \bot
\\[0ex] \bot & \bot & \top & \top & \top & \top \\ \hline
\end{array}$$
We can then verify that the last two columns are in agreement on every truth assignment.
$(B⇒C)∧(C⇒B)$ is true if and only if $B$ and $C$ have the same truth value.
Yes.
Thus, $B$ and $C$ are logically equivalent.
Not quite. It's "Thus, $B$ and $C$ are logically equivalent if and only if $B$ logically implies $C$ and $C$ logically implies $B$".
Which was to be demonstrated.
$\Box$
Best Answer
Actually, they both have the same truth table. They only look different because you omitted $R$ from the second one, presumably due to the fact that it does not appear in the second formula. If we include $R$ in the second truth table, it becomes
As you can see, this is the same as the first one.