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Proof step 49 from proof diff-primes (previous step)

abs(n→i(m)) < j, n→i(m) = n→i(n) % j, n→i(m) + n→i(n) < j, 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, 1 < j, 0 < j, 0 < 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 50:
n→i(m) = n→i(n) % j, n→i(m) < j, 0 < n→i(m) + n→i(n), 0 < j, 1 < 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, n→i(m) + n→i(n) < j ⊦ ~1 = (j - 1) / 2, m = n, j = 2, countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = []

Rule argument

Additional information

Used simplifier rules: mod-less-01: ⊦ 0 < j → i % j < j; from specification int-div%lel: ⊦ i < i0 ∧ i = i1 → i1 < i0; (forward) from specification oelem[intlista]%abs-02: ⊦ abs(n→i(n)) = n→i(n); from specification intnat