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Proof step 5 from proof even-not-ex (previous step)

i < 0 ⊦ i % 2 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1)

User applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0)
Proof step 7:
i = i1 * 2 + j0, i < 0, abs(j0) < 2, 0 < 2, j0 ≤ 0, i1 ≤ 0 ⊦ j0 = 0 ↔ ¬(∃ i0. i = 2 * i0 + 1)

Rule argument

spec: <toplevel> name: div-elim-negpos seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j ∧ i0 ≤ 0) subst: [i, j, i0, j0] → [i, 2, i1, j]

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: i, 2, i1, j