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Proof step 24 from proof div-ls-mone (previous step)

0 < j, i / j < ~1, i / j < 0, 0 ≤ i ⊦ i + 2 * j ≤ 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 25:
⊦ 0 < j ∧ 0 ≤ i → (i0 = i / j ∧ j0 = i % j ↔ i = i0 * j + j0 ∧ 0 ≤ j0 ∧ j0 < j ∧ 0 ≤ i0)

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 simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i ≤ j ↔ ¬ j < i; (ss) Used terms: i, j, i0, j0