The set of all functions from A to B is denoted AB. Prove that NP(N)∼P(N).
Previous question proved that for any set A, A{yes,no}∼P(A). The symbol ∼ means equinumerous to. N does not include 0 here. P is power set operation. I know we have to create a bijection between NP(N)∼P(N). I believe I might be close to a solution, but am looking for some suggestions first.
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