# Is concatenation of secure encryptions secure?

If I have $$n$$ secure assymetric encryption schemes, is their concatenation $$Enc_{k_1,...,k_n}(m)=Enc_{k_1}(m)||...||Enc_{k_n}(m)$$ secure?

My intuition is that the concatenation is secure, but I can't solve it by reduction, which gives me the feeling that maybe an attacker can somehow take advantage of the combination of the encryptions.

The main security definition I'm interested in is the following - an encryption scheme is secure if for all PPT $$A$$ and function that maps public keys to messages $$f$$, the probability of $$A$$ to guess a random bit $$b$$ chosen uniformly when it sees $$p_k,f(p_k)=(m_0,m_1),Enc_{pk}(m_b)$$ is at most $$\frac{1}{2}+negl(n)$$, where the probability is over the coin tosses of $$A$$ and the public key $$p_k$$ generation.

• This is an easy question if you consider how strong a chain... Commented Mar 17 at 21:14
• @MaartenBodewes-modelection - thanks for your comment. This is non an homework question, that's a question I thought about and couldn't solve. I didn't think my thoughts about it are so helpful, so I just posted the question. I also think that the existence of this question with answer in this site would help others. Anyway, I edited the question to include some of my thoughts. Commented Mar 18 at 6:30
• @kelalaka - What do you mean by chain? And what is its strength? Commented Mar 18 at 6:31
• That depends entirely on the definition of security. Commented Mar 18 at 6:42
• @Maeher - Can you elaborate? I also added security definition. Commented Mar 18 at 6:48

Consider the following proof by contradiction: if an adversary would receive $$\text{Enc}_{k_1}(m)$$. Then the adversary would be able to try and guess $$m$$ and construct $$\text{Enc}_{k_1,...,k_n}(m)=\text{Enc}_{k_1}(m)||...||\text{Enc}_{k_n}(m)$$ using the public keys of locally generated key pairs. If that would leak any information then the initial cipher would not be IND-CPA secure. This in turn would mean that no public key encryption schemes would be secure.