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Proof step 46 from proof diff-primes-neg (previous step)

0 < j ∧ 0 ≤ n→i(n) ∧ 0 ≤ n→i(m), n→i(n) + n→i(m) < 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, n→i(n) % j = n→i(m) % j, 1 < j, 0 < j, 0 < n→i(n) + n→i(m) ⊦ 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 47:
n→i(n) % j = n→i(m) % j, n→i(n) + n→i(m) < 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(n) + n→i(m) ⊦ 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: nat2int: ⊦ 0 ≤ n→i(n); from specification intnat