Exercise A.1.4
Suppose that you have shown that whenever X is true, then Y is true, and whenever Y is false, then X is false. Have you now demonstrated that X is true if and only if Y is true? Explain.
This is similar to the previous exercise A.1.3 and the approach is the same, to enumerate the truth table for the given statements.
The following is the truth table for the first statement "whenever X is true, then Y is true", which is the implication $X \implies Y$.
| X | Y | X $\implies$ Y |
|---|---|---|
| F | F | T |
| F | T | T |
| T | F | F |
| T | T | T |
The following truth table is for the second statement "whenever Y is false, then X is false", which is the implication $\neg Y \implies \neg X$.
| X | Y | $\neg Y$ | $\neg X$ | $\neg Y \implies \neg X$ |
|---|---|---|---|---|
| F | F | T | T | T |
| F | T | F | T | T |
| T | F | T | F | F |
| T | T | F | F | T |
We can see the truth values for $X \implies Y$ and its contrapositive $\neg Y \implies \neg X$ are the same. It is a well known fact that an implication and its contrapositive are logically equivalent.
Let's now consider the conjunction of these two statements.
| X | Y | $X \implies Y$ | $\neg Y \implies \neg X$ | $(X \implies Y) \land (\neg Y \implies \neg X)$ |
|---|---|---|---|---|
| F | F | T | T | T |
| F | T | T | T | T |
| T | F | F | F | F |
| T | T | T | T | T |
Unsurprisingly the truth values of the conjunction are the same as the conjuncts.
Now to answer the question. The truth values are not the same as for logical equivalence, $\iff$.
So, no, we have not demonstrated that X and Y are logically equivaelent. $\square$
(Thanks to Issa Rice for advising the original answer was misguided.)
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