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Proof step 19 from proof mod-mult-add (previous step)

i0 < j, i0 ≤ 0, 0 < j, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ i1 ≤ 0 ∧ i1 < j → (i + i1 * j) |%| j = i |%| j ⊦ 0 < j ∧ i0 + 1 ≤ 0 ∧ i0 + 1 < j, (i + i0 * j) |%| j = i |%| j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 20:
i0 < j, i0 ≤ 0, 0 < j, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ i1 ≤ 0 ∧ i1 < j → (i + i1 * j) |%| j = i |%| j ⊦ (i + i0 * j) |%| j = i |%| j, i0 < 0 ∧ i0 + 1 < j

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%mone-05: ⊦ i ≤ ~1 ↔ i < 0; from specification int-abs%numle-03: ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j; from specification int-abs