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We prove that every natural number greater than 1 can be factorized into a sequence of prime numbers, i.e.,
We proceed by complete induction over n.
forall n in N: n > 1 => (exists k in N, f: Nk -> N: n = (prod0 <= i < k f(i)) /\ forall i in Nk: f(i) is prime).
We take arbitrary n in N and assume
We have to show
(1) forall m < n: m > 1 => (exists k in N, f: Nk -> N: m = (prod0 <= i < k f(i)) /\ forall i in Nk: f(i) is prime).
n > 1 => (exists k in N, f: Nk -> N: n = (prod0 <= i < k f(i)) /\ forall i in Nk: f(i) is prime).
See lecture notes.