Here is the best I'm able to do so far, following up on my idea in the comments:

Let $P(x)$ be the claim that any set of $x$ integers can be ordered in a sequence $s_i$ with $1 \le i \le x$ such that if $a \lt b \lt c$, then $s_a + s_b \ne s_c$.

Suppose $S$ is a set containing $2 \cdot x$ integers. We have $S = V \cup W$ where $V$ contains the $x$ smallest elements of $S$ and $W$ contains the $x$ largest elements of $S$. If $V$ and $W$ each correspond to a sequence with the property then we can make a sequence for $S$ by starting with $W$'s sequence and appending $V$'s sequence to it. So $\forall x P(x) \implies P(2 \cdot x)$.

If a set can be ordered in such a way so can all of its subsets: to find a sequence for a set after having removed some elements, simply remove those same elements from the original set's sequence. So if $y < x$ then $P(x) \implies P(y)$, and in particular this gives $\forall x P(3 \cdot x + 2) \implies P(2 \cdot x + 1)$.

Therefore if the Collatz conjecture is true, $\forall x P(x)$.

Now this is disappointing because it is just a long-winded way of saying sort the set from highest to lowest, also it proves much more than $P(3 \cdot x + 2) \implies P(2 \cdot x + 1)$. Maybe a starting point for a better example.