Complete Induction

Content:
[∀ n ∈ N, p(0), ..., p(n - 1) => p(n)] => ∀ n ∈ N, p(n)

Example:
Claim: Every natural number greater than 1 has a prime factorization.
Proof:
Assume that ∀ n ∈ N - {0, 1} and that for all natural numbers 1 < j < n, p(j) is true. -> IH
Then n is either prime or not prime.
Case1: n is prime Then n = n is its own prim factorization.
Case2: n is not prime Then n has ate least 3 factors, 1, n, and x ≠ 1, x ≠ n. So 1 < x < n, som x is a factor different from 1 and n. Then also, n has a factor y = n/x, y ≠ n and y ≠ 1, since otherwise xy ≠ n.
Now by IH x and y have prime factorization, we conclude n has a prime factorization, that is p(n).
Since n was arbitrary, this shows that ∀ n ∈ N - {0, 1}, [p(2), p(3), ..., p(n - 1)] => p(n)
Conclude ∀ n ∈ N - {0, 1},  n has a prime factorization.