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Proof step 8 from proof div-two-intro (previous step)

⊦ i % 2 = 0, i / (j0 * 2) = (i / 2) / j0, j0 = 0

User applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 10:
⊦ 0 < 2 ∧ 0 ≤ i, i % 2 = 0, i / (j0 * 2) = (i / 2) / j0, j0 = 0
Proof step 27:
i = i0 * 2 + j1 ∧ 0 ≤ j1 ∧ j1 < 2 ∧ 0 ≤ i0 ∧ 0 < 2 ∧ 0 ≤ i ⊦ j1 = 0, i / (j0 * 2) = i0 / j0, j0 = 0

Rule argument

spec: int-div inst: name: div-elim-pospos seq: ⊦ 0 < j ∧ 0 ≤ i → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0) subst: [i, j, i0, j0] → [i, 2, i0, j] flags: rewrite, current

Additional information

Used terms: i, 2, i0, j