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Proof step 3 from proof div-shift-two (previous step)

0 < j, 0 ≤ i ⊦ i / j = i / (2 * j) * 2 + (i < j * (i / (2 * j) * 2 + 1) ⊃ 0;1), j = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)
Proof step 5:
i = i0 * j + j0, 0 < j, j0 < j, 0 ≤ i, 0 ≤ j0, 0 ≤ i0 ⊦ i0 = i / (2 * j) * 2 + (i < j * (i / (2 * j) * 2 + 1) ⊃ 0;1), j = 0

Rule argument

spec: <toplevel> 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, i0, j0] flags: <rotated> <all paths>

Additional information

Used terms: i, j, i0, j0