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I am looking to use randomized public-key encryption in a context where it should also serve as a sort of "binding commitment". That is, I want to encrypt a value $x$ with some randomness $rnd$ under a public key $pk$, denoted $enc(x, rnd, pk)$ for short, such that it is infeasible to find a different plaintext $x'$ and randomness $rnd'$ such that $enc(x, rnd, pk) = enc(x', rnd', pk)$ (the ciphertexts are the same).

I want this because a party A should be able to convince a party B that a given ciphertext $c$ is an encryption of a plaintext (known to the parties) $x$. For this, A would send the ciphertext $c$ and the randomness $rnd$ to B, and B would re-compute $enc(x, rnd, pk)$. When this equals $c$, then B should be convinced that $c$ is an encryption of $x$.

Is this a property that randomized public-key encryption schemes usually have, or can be derived from another property?

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Yes, this is a property that most randomized public-key encryption schemes have. In fact, every standard public-key encryption scheme you could think of satisfies this property (e.g. RSA-OAEP, ElGamal, Goldwasser-Micali, Paillier...).

However, the binding property is not implied by the semantic security of an encryption scheme. In fact, under very strong cryptographic assumptions (the existence of a primitive called indistinguishability obfuscation), there exists public-key encryption schemes which are deniable, in the sense that for any ciphertext $c$ and any plaintext $m$ (possibly different from the one encrypted in $c$), a party can always find a randomness $r$ such that $c = \mathsf{Enc}(m;r)$ (see this paper). Such encryption schemes are interesting objects (intuitively, they would allow to guarantee security even if the government break into your house, point a gun on your head, and asks you to reveal the random coins that you used to encrypt a message - with a deniable encryption scheme, you will always be able to lie and to claim that you encrypted an innocent message). By definition, such encryption scheme are semantically secure but not binding.

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  • $\begingroup$ Thank you for your answer! While this answers my question, I do have a small follow-up: I have trouble finding formal proofs for statements such as the first one you made (that these public-key encryption schemes have this property). Since I am interested in using it in a protocol for which I would like formal proofs, and this property is needed for the security of my protocol, I was wondering if I could find something of the sort for a concrete scheme (given some mathematical hardness assumptions). $\endgroup$ – Sven Hammann Apr 11 '18 at 8:36
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    $\begingroup$ For all standard schemes, you will not even need any mathematical assumption: they are perfectly binding. Take for example ElGamal: given a public key $(g,h)$ and a ciphertext $(c_0,c_1)$, there is a unique pair $(M,r)$ such that $(c_0,c_1) = (g^r, h^rM)$. Therefore, the property that you want is perfectly guaranteed, without any assumption: it is completely impossible to find two different pairs that lead to the same ciphertext. The same is true for RSA, Paillier, etc. And proving this simply amounts to verifying that a ciphertext uniquely defines a pair (message, random coin). $\endgroup$ – Geoffroy Couteau Apr 11 '18 at 11:45

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