Thanks for engaging in this discussion and explaining your thoughts. I'm still trying to understand.

> given ciphertext C = A ^ B you can decrypt plaintext A using key B

Right, C ^ B = A, to get back the original.

Already here, it seems to me that C is not pure random noise if A can be recovered from it. C contains the encoded information of A, so it's not random. Is that wrong?

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> or you can decrypt plaintext D using key E = C ^ D

In this example, isn't E the cyphertext and C is the key? E (cyphertext) = D (original) ^ C (key).

Then E ^ C = D, to get back the original.

Here, it seems to me that E contains the encoded information of D, so it's not random. And C plays the role of a key similar to B, so it's not being decoded as a ciphertext in this case, and nothing is implied whether it's random or not.

> XOR is commutative so for a one-time pad key and ciphertext can be swapped.

Maybe this is what I'm missing. In the second example, I'll swap the ciphertext and key:

C ^ E = D

Hmm, so both C and E are required to recover D the original information. Does that mean that the information was somehow distributed into C and E, so that they are both meaningful data, not random?

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But what about B in the first example, a key that was supposed to be pure random? If I swap the ciphertext and key:

B ^ C = A

The information in A is recovered from the interaction of both B and C. So the entirety of the information is not in C, the ciphertext (or key in this case).

Does that mean, B is no longer pure noise? When it was used as a key, it became meaningful in relation to A. That's information, isn't it?