# Dynamic accumulator with only non-membership witness

Goal: Construct a dynamic accumulator which can add elements and provide non-membership witnesses. I wanted to know if my construction is correct.

Consider a group $$Z_p$$ where $$p$$ is prime number. Consider the set $$S=\{c_1, c_2,\cdots, c_m\}$$ where $$c_i\in Primes(Z_p)$$. Consider the following accumulator: $$A = \left(\prod_{c_i \in S} c_i\right) \text{mod } p.$$

Addition: To add a new element $$c \in Primes(Z_p)$$, $$A \longleftarrow \left(A\times c\right) \text{mod } p.$$

Non-Memerbship witness $$(x,y)$$: To show $$c \not\in S$$, we use the fact that $$c$$ is co-prime with all the elements in $$S$$. We use Bezout coefficient $$(x,y)$$ to show that $$cx+Ay=1 \text{ mod } p.$$

Security: $$\forall c\in S$$, there exists no algorithm $$\mathcal{A}$$ such that

$$(x,y) \leftarrow \mathcal{A}(c,S),\; cx+Ay=1 \text{ mod } p.$$

• See the Construction 2 of lilixue paper for nonmembership-witness Jul 14, 2019 at 20:33
• @Narodism Welcome to crypto.stackexchange - It would help improve the quality and reception of your answer if you could link to the relevant paper and summarize the relevant portion of it here Jul 14, 2019 at 23:02
• @Narodism I have converted your answer to a comment for now as it seems more like a reference suggestion than a proper answer. Please use Ella's advice to write a new answer once you have the summary. Jul 15, 2019 at 13:05

Your scheme assumes that the following holds: if $$a,b$$ are primes and $$0< a,b < p$$, then $$(a,b)=1 \implies (ab\mod p,a)\neq1.$$ If this is true, then your scheme is correct, however this is not the case.
Toy example: $$p=11,a=5,b=7$$. Since $$ab \mod 11=35 \mod 11=2,$$ therefore $$A=2$$. Now, it is clear that security does not hold, since one could prove false non-membership statements. The reason for that is: $$(a,A)=1=(b,A)$$. Meaning one can prove that $$a$$ and $$b$$ are not accumulated however they were indeed added to the accumulator. For instance $$(2,1)$$ is a a non-membership witness for $$5$$, when $$A=2$$, since $$5*2+2*1 \mod 11 = 1$$.