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Proof step 24 from proof diff-primes-neg (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
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(n) ^2 - n→i(m) ^2, 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

Additional information

Used simplifier rules: forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%div-abs-02: ⊦ i | abs(j) ↔ i | j; from specification int-gcd%ax-04: ⊦ i + 0 = i; from specification int-basic2%sub-sub-04: ⊦ i - (j - j0) = i + j0 - 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%minus-add: i < 0 ⊦ (* i) + j = j - ~ i; from specification int-abs%sub-add-01: ⊦ i + (j - i0) = i + j - i0; from specification int-abs%sub-sub: ⊦ i - j - i0 = i - j + i0; from specification int-abs