Exercise A.1.2
What is the negation of the statement “X is true if and only if Y is true”? (There may be multiple ways to phrase this negation.)
This English statement can be interpreted in several ways if read informally.
To a mathematician, the phrase "if and only if" has a precise meaning. It is the bi-conditional $X \iff Y$, which is the combination of two implications $(X \implies Y) \land (Y \implies X)$.
Let's try to negate the statement at an English level:
- "it is not the case that X is true if and only if Y is true"
- .. leads to "X is true if Y is false, and, Y is true if X is false"
It is not immediately apparent that this also means the statement is false when X and Y are both true, or both false. It needs some thinking. Due to this risk, it is good practice to write out the truth tables.
X | Y | X $\implies$ Y |
Y $\implies$ X | X $\iff$ Y | $\neg$ (X $\iff$ Y) |
---|---|---|---|---|---|
F | F | T | T | T | F |
F | T | T | F | F | T |
T | F | F | T | F | T |
T | T | T | T | T | F |
Here we've used the truth value definition of implication, which itself is counter-intuitive and requires a separate blog post to discuss.
The truth table confirms the nation of the given statement is "at most one of X or Y is true", or "either X or Y is true, but not both", or simply the exclusive-or relation $(X \oplus Y)$. $\square$
Note that combining this result and the one from the last exercise A.1.1, we can see XOR and logical equivalence are negations of each other.
$$ \boxed{ \neg (X \oplus Y) = (X \iff Y) }$$
No comments:
Post a Comment