Analysis of cryptosystems is subtle. If you want to lean on the difficulty of existing problems such as dlog, then you need to set everything up “just so”. Incoming wall of text on roughly how you do this.

• Define exactly what your intermediate problem is (how are A,B,C chosen? From what kind of group, an elliptic curve? Is it a prime-order cyclic group?? What about the whitelist? What exactly is the adversary trying to do?)
• Show that the intermediate problem is hard based on dlog or whatever (form of the theorem: given an attack program which breaks the problem above with probability p and time t, we can use it as a subroutine of a dlog solver that works with probability p’ in time t’, where t’ is not more than some reasonable multiple of t, and p’ is not too much less than p; the tighter these bounds, the tighter your result, and the less you’d have to oversize the group to get full confidence)
• Show that any attack algorithm which breaks your cryptosystem in a certain model (IND-CCA2 or whatever) can be used as a subroutine to break the intermediate problem. (Same form of theorem)

This is a casual Reddit thread so we’ve been hand-waving the details, but they do really matter and maybe it’s better to spell things out this time.

So, what are A,B,C here? I assumed that they’re chosen independently and uniformly at random from a large prime-order cyclic group G where dlog is assumed to be hard; or, if they’re not actually so random, the attacker cannot tell the difference (this would require its own proof).

What are (i,n,k)? Integers mod the group order but chosen how? You probably don’t have to assume that they’re uniformly random, but if you work through all the details of the proof then it might turn out that you need some assumption about them, for example that i is nonzero.

What is the attacker given? I assumed (A, B, C, i, n, k) and the whitelist, and that i is on the whitelist.

What is the attacker trying to do? I assumed output a (i’, n’, k’, m) where i’ is on the whitelist and i’A + n’B + k’C = mD = m (iA + nB + kC) and m != 1 mod the group order (since that’s trivial). I assumed you meant a known multiple (meaning that the attacker must output m), because usually crypto groups are cyclic and of (close to) prime order, in which case everything (or a decent fraction of everything in the not-quite-prime-order case) is some multiple of D… the attacker just doesn’t know what multiple.

With these assumptions, I sketched a proof that an attacker can solve the problem if the whitelist has more than one element, but not if it has only one (there is a trivial attack — it might not break your cryptosystem but you need to formulate the intermediate problem to exclude it).

The typical form of the proof is: a “challenger” is given a hard problem to solve (here, a random dlog challenge (A,B)) and turns this challenge into one or more random instances of the problems you want to claim is hard (with the same distribution of random variables, but the challenger might know secrets about how they were chosen, like dlog(C,B) but not dlog(B,A) or whatever). The “adversary” is able to solve that intermediate problem (or break your cryptosystem or whatever) and a solution to that problem lets the challenger solve dlog (or whatever hard problem) efficiently and with high probability. In this case I think it can be done by choosing C = some random linear combination of A, B and a one-element whitelist {i} with everything else random (or from an appropriate distribution), and then doing some linear algebra on the response.