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Proof step 2 from proof div-ls (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
i0 % j = 0, i < i0, 0 < i0, 0 < j, j0 ≤ i / j, i0 / j ≤ j0, 0 ≤ i, i0 / j ≤ i / j ⊦

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%forward-lele: ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)