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Proof step 29 from proof diff-primes (previous step)

prime(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, j | n→i(m) - n→i(n) ∨ j | n→i(m) + n→i(n), 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. 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 30:
⊦ i < j ∧ 0 < i → ¬ j | i
Proof step 31:
¬(n→i(m) + n→i(n) < j ∧ 0 < n→i(m) + n→i(n)), prime(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, j | n→i(m) - n→i(n) ∨ j | n→i(m) + n→i(n), 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2
Proof step 43:
n→i(m) + n→i(n) < j ∧ 0 < n→i(m) + n→i(n), prime(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, j | n→i(m) - n→i(n) ∨ false, 0 < j ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

spec: int-gcd inst: name: div-not-ls seq: ⊦ i < j ∧ 0 < i → ¬ j | i subst: [i, j] → [n→i(m) + n→i(n), j] flags: rewrite

Additional information

Used terms: n→i(m) + n→i(n), j