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Proof step 14 from proof diff-primes-neg (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ n→i(n) ≤ j - i ∧ 0 ≤ i ∧ i ≤ j → countFun(i, j, intfun)[n] = intfun(i + n→i(n))
Proof step 16:
¬(n→i(n) ≤ (j - 1) / 2 - 0 ∧ 0 ≤ 0 ∧ 0 ≤ (j - 1) / 2), (~1 - n→i(m) ^2) |%| j = countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)[n], 0 ≤ j, m < i→n(1 + (j - 1) / 2), n < i→n(1 + (j - 1) / 2), prime(j), n→i(m) ≤ (j - 1) / 2 ⊦ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2
Proof step 19:
n→i(n) ≤ (j - 1) / 2 - 0 ∧ 0 ≤ 0 ∧ 0 ≤ (j - 1) / 2, (~1 - n→i(m) ^2) |%| j = (~1 - (0 + n→i(n))^2) |%| 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 ⊦ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

spec: intlist inst: name: countFun-at seq: ⊦ n→i(n) ≤ j - i ∧ 0 ≤ i ∧ i ≤ j → countFun(i, j, intfun)[n] = intfun(i + n→i(n)) subst: [n, intfun, j, i] → [n, λ i. (~1 - i ^2) |%| j, (j - 1) / 2, 0] flags: rewrite

Additional information

Used terms: n, λ i. (~1 - i ^2) |%| j, (j - 1) / 2, 0