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Proof step 20 from proof diff-primes (previous step)

n→i(m) ^2 |%| j = 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, n→i(n) ≤ (j - 1) / 2 ⊦ 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 21:
⊦ 0 < j → (≡mod(i, i0, j) ↔ i |%| j = i0 |%| j)
Proof step 22:
0 < j, ≡mod(n→i(m) ^2, 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, n→i(n) ≤ (j - 1) / 2 ⊦ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) = [], j = 2, m = n, ~1 = (j - 1) / 2

Rule argument

spec: inst: name: congmod-def-mod seq: ⊦ 0 < j → (≡mod(i, i0, j) ↔ i |%| j = i0 |%| j) subst: [i0, j, i] → [n→i(n) ^2, j, n→i(m) ^2] flags: rewrite, current

Additional information

Used simplifier rules: lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%anti: ⊦ i ≤ j → (j ≤ i ↔ i = j); from specification int-abs%countFun-one: ⊦ countFun(i, i, intfun) = intfun(i) '; from specification intlist%zero-i: ⊦ 0 ^2 = 0; from specification int-square%e: ⊦ i ' ≠ []; from specification list[intlista]%nat2int-10: ⊦ n→i(n) ≤ 0 ↔ n = 0; from specification intnat Used terms: n→i(n) ^2, j, n→i(m) ^2