# Does absence of second pre-image resistance infer absence of pre-image resistance?

Talking about hash functions, I know the hierarchy is as follows:

1. Pre-image resistance
2. Second pre-image resistance
3. Collision resistance

Where each property implies the one before it so 2nd pre-image resistance implies pre-image resistance.. But is this also true the other way around?

Can we infer from the absence of second pre-image resistance, that there would also be the absence of pre-image resistance?

$$H(x) = SHA512(Trunc(x))$$
where $Trunc(x)$ just returns $x$ with the last byte removed.
$H$ is not second-preimage resistant; given $H(x)$, we can change the last byte of $x$ to another value that hashes to the same value. However, given a target hash value $z$, we can't find a value $x$ with $z = H(x)$; if we did, then we could find preimages to SHA512.