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Proof step 2 from proof div2-nle-02 (previous step)

0 < j, i < (i + j) / 2, 0 ≤ i, i < j, j ≤ (i + j) / 2, 0 < (i + j) / 2 ⊦

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:
i + j = i0 * 2 + j0 ∧ 0 ≤ j0 ∧ j0 < 2 ∧ 0 ≤ i0 ∧ 0 < 2 ∧ 0 ≤ i + j, 0 < j, i < i0, 0 ≤ i, i < j, j ≤ i0, 0 < i0 ⊦

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, 2, i0, j0] flags: rewrite, current

Additional information

Used simplifier rules: lstwozero: ⊦ i < 0 → ¬ 0 < i / 2;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs Used terms: i + j, 2, i0, j0