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Proof step 52 from proof diff-primes (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 53:
0 ≤ j, m < i→n(1 + (j - 1) / 2), n < i→n(1 + (j - 1) / 2), prime(j), n→i(m) ≤ (j - 1) / 2, n→i(n) ≤ (j - 1) / 2, j | n→i(m) ^2 - n→i(n) ^2, 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2, n→i(m) ^2 = (n→i(m) - n→i(n)) * (n→i(m) + n→i(n)) + n→i(n) ^2

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs