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Proof step 59 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) * n→i(m) - n→i(n) * n→i(n), 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2, n→i(m) * n→i(m) = (n→i(m) * (n→i(m) + n→i(n)) - n→i(n) * (n→i(m) + n→i(n))) + n→i(n) * n→i(n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
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) * n→i(m) - n→i(n) * n→i(n), 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, ~1 = (j - 1) / 2, n→i(n) * n→i(n) + n→i(m) * (n→i(m) + n→i(n)) = n→i(m) * n→i(m) + n→i(n) * (n→i(m) + n→i(n))

Rule argument

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs