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Proof step 32 from proof diff-primes (previous step)

0 < j, n→i(n) ≤ (j - 1) / 2, n→i(m) ≤ (j - 1) / 2, prime(j), n < i→n(1 + (j - 1) / 2), m < i→n(1 + (j - 1) / 2), 0 ≤ j, j | n→i(m) - n→i(n) ∨ j | n→i(m) + n→i(n) ⊦ ~1 = (j - 1) / 2, m = n, j = 2, countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], n→i(m) + n→i(n) < j ∧ 0 < n→i(m) + n→i(n)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 5

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