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Proof step 13 from proof lagrange-lemma (previous step)

prime(j) ⊦ j = 2, filter(countFun(0, (j - 1) / 2, λ i. i ^2 |%| j), countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j)) = [], ∃ i, i0, j0. 1 + i ^2 + i0 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ filter(ints, ints0) = [] ↔ (∀ i. i ∈ ints → ¬ i ∈ ints0)
Proof step 15:
prime(j) ⊦ j = 2, ∀ i. i ∈ countFun(0, (j - 1) / 2, λ i. i ^2 |%| j) → ¬ i ∈ countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j), ∃ i, i0, j0. 1 + i ^2 + i0 ^2 = j0 * j ∧ 0 < j0 ∧ j0 < j

Rule argument

spec: list-set inst: intlista name: filter-empty-def seq: ⊦ filter(ints, ints0) = [] ↔ (∀ i. i ∈ ints → ¬ i ∈ ints0) subst: [ints0, ints] → [countFun(0, (j - 1) / 2, λ i. (~1 - i ^2) |%| j), countFun(0, (j - 1) / 2, λ i. i ^2 |%| j)] flags: rewrite

Additional information

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