Can someone help me understand this logic problem:
What are the truth values of these statements?
a) $\exists !\,x\;P(x) \implies \exists x\;P(x)$
b) $\forall x\;P(x) \implies\exists !x\;P(x)$
c) $\exists!x\;\neg P(x) \implies \neg\forall x\;P(x)$
Correct answer (from the book answers):
The way I am see a) for example is:
- "There is a unique x for which if $P(x)$ then there exists another $x$ for which $P(x)$"
- $\exists !\,x\;P(x)$ can be broken down to just $P(x)$ since there is a "unique $x$"
- $\exists x\;P(x)$ can be broken down to: $P(x_1) \lor P(x_2) \lor ... \lor P(x_n)$
- Therefore at step #2, if the unique $x$ is FALSE, then the entire statement is TRUE (according to the truth table for $\implies$)
- If #2 is TRUE, then step #3 must have at list 1 TRUE $P(x_i)$ for the overall statement to be true,
- Therefore I see it can be both TRUE and FALSE depending on $x$. I am not sure how the book comes to true. I don't fully understand this. Can someone correct me and fill my gap in understanding there. Similar issues in understanding b) and c) as well...