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Proof step 2 from proof modeq-pos-03 (previous step)

j0 < i0, 0 < i0, j0 < j, j0 + 1 < j, 0 ≤ i0 * j + j0, 0 < j, j0 < 0, j0 + 1 < i0 ⊦ (i0 * j + j0) % j = j0 % j + j, j0 % j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
⊦ 0 < j ∧ i < 0 → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j)
Proof step 4:
j0 = i1 * j + j1 ∧ j1 ≤ 0 ∧ abs(j1) < j, j0 < i0, 0 < i0, j0 < j, j0 + 1 < j, 0 ≤ i0 * j + j0, 0 < j, j0 < 0, j0 + 1 < i0 ⊦ (i0 * j + j0) % j = j1 + j, j1 = 0

Rule argument

spec: int-div inst: name: div-elim2 seq: ⊦ 0 < j ∧ i < 0 → ( i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ j0 ≤ 0 ∧ abs(j0) < j) subst: [i, j, i0, j0] → [j0, j, i, j1] flags: rewrite, current

Additional information

Used terms: j0, j, i, j1