In Zeroth-Order Logic, a formula is said to be consistent if it is TRUE under at least one valuation.
So I am checking if $ P \implies \lnot P $. Here is how it looks like:
For $ P = \text{TRUE}, $ we have $ P \implies \lnot P = \text{FALSE}.$
For $ P = \text{FALSE}, $ we have $ P \implies \lnot P = \text{TRUE}.$
So can we say that $ P \implies \lnot P $ is consistent?
If we can say this is consistent, then it somehow defies intuition, does it not? A proposition that implies the negation of it is impossible. Can you think of such a proposition? What am I missing?
Yes. In fact, not only is $P$ being false a valuation that sets this statement to True, but you can easily show that this statement is logically equivalent to $\neg P$:
$$P \to \neg P \overset{Implication}{\Leftrightarrow} \neg P \lor \neg P \overset{Idempotence}{\Leftrightarrow} \neg P$$
I believe the feeling that it defies intuition that this statement would be consistent stems from the fact that it feels there is some contradiction within thois statement ... and indeed: if you assume that $P$ is True, then that would mean that $\neg P$ is True as well, and hence the contradiction. However, all that that really means is that $P$ apparently cannot be True. Hence, it needs to be False. So, what this statement implies is that $P$ has to be False, i.e. $\neg P$. And as it turns out, it is in fact equivalent to it.