Security of Diffie-Hellman with multiplication for secret derivation?
This problem (in the computational sense) is equivalent to the CDH problem [1].
This problem can be summarized as &quot;given $g, A, A^y, B, B^x$, compute $g^x \cdot g^y$
Given a CDH oracle, we can solve this with two queries; we get the inputs $g, A, A^y, B, B^x$, and we give the CDH oracle the values $B, B^x, g$, and it'll give us the value $g^x$; we do the same to get $g^y$, and a simple multiplication gives us $g^x \cdot g^y$
Conversely, given an oracle that solves this problem, we can solve the CDH problem; we get the inputs $C, C^x, D$, we select an arbitrary $E, y$ values, and give the oracle the tuple $D, E, E^y, C, C^x$; it'll return the value $D^x \cdot D^y$.  We then multiply that by $D^{-y}$, and that gives us the answer to the CDH problem $D^x$

Now, addressing the decisional problem would appear to be trickier; it's easy enough to show how to solve the DDH problem with an oracle that solves the decisional version of this problem; however, I don't see an obvious way to use a DDH oracle to solve this problem.

[1]: Note: for the CDH problem, I won't be using the usual formulation &quot;given $g, g^x, g^y$, return $g^{xy}$, instead, I'll be using the equivalent &quot;given $g, g^x, h$ return $h^x$.  That formulation would appear to be make the proof a bit cleaner...