Given a set of conditions listed below:
For some predicate $P(n), P(1)$ and $P(2)$ are True, but $[(\forall n ≥ 3)(P(n) \implies P(n + 1))]$ is False Does this predicate exist or not? Give an example or provide reasoning.
I feel as if this couldn't exist. For all $n$ greater or equal to three, $P(n) \implies P(n+1)$ to be false, $P(n)$ must be true and $P(n+1)$ must be false. Given an arbitrary $m$, where $m = n+1$, then for the same statement $P(m) \implies P(m+1)$, then $P(m)$ must be true and $P(m+1)$ must be false. But we've already stated that $P(m) = P(k+1)$ which must be false, creating a contradiction.
Likewise, another example would be $P(1)$ is True, $P(k) \implies P(k + 1)$ is False for all $k \in \mathbb N$.
Apologies for formatting issues, I am still new to this site :)
$\lnot \forall n{\in}\Bbb N_{\geq 3}~(P(n)\to P(n+1))$ is not equivalent to $\forall n{\in}\Bbb N_{\geq 3}~\lnot(P(n)\to P(n+1))$.
You just require $\exists n{\in}\Bbb N_{\geq 3}~(P(n)\land\lnot P(n+1))$
So $P(0),P(1),P(2),P(3),\lnot P(4),\ldots$ will suffice.