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Proof step 2 from proof div-le-pos (previous step)

j0 ≤ i / j, i ≤ i0, i0 / j < j0, 0 < j, i0 / j < i / j ⊦

User applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 4:
j0 ≤ i / j, i ≤ i0, i0 / j < j0, 0 < j, i0 / j < i / j ⊦ 0 < j ∧ 0 ≤ i
Proof step 18:
i = i1 * j + j1 ∧ 0 ≤ j1 ∧ j1 < j ∧ 0 ≤ i1 ∧ 0 ≤ i, j0 ≤ i1, i ≤ i0, i0 / j < j0, 0 < j, i0 / j < i1 ⊦

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, j, i1, j1] flags: rewrite, current

Additional information

Used terms: i, j, i1, j1