It's exercise 1.8 in Introduction to Mathematical Logic, Sixth Edition by Elliot Mendelson.
I couldn't prove it with the definitions given by the author. How should I approach the problem?
A is said to logically imply B if and only if every truth assignment to the statement letters of A and B that makes A true also makes B true.
A is said to be logically equivalent to B if and only if A and B receive the same truth value under every assignment of truth values to the statement letters of A and B. For example, $A$ and $\neg \neg A$ are logically equivalent.
Below is my proof that feels incorrect.
By proposition 1.1.a, $B \Rightarrow C$ and $C \Rightarrow B$ are tautologies.
Since $B \Rightarrow C$ and $C \Rightarrow B$, $(B \Rightarrow C) \wedge (C
\Rightarrow B)$.
\begin{array}{lllll}
B & C & B \Rightarrow C & C \Rightarrow B & (B \Rightarrow C) \wedge (C
\Rightarrow B)\\
T & T & T & T & T\\
T & F & F & T & F\\
F & T & T & F & F\\
F & F & T & T & T
\end{array}
$(B \Rightarrow C) \wedge (C \Rightarrow B)$ is true if and only if $B$ and
$C$ have the same truth value. Thus, $B$ and $C$ are logically equivalent.
Best Answer
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.
Yes.
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.
$\Box$