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Proof step 10 from proof mod-mult-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
(i + (i0 - 1) * j) |%| j = i |%| j, 0 < j, 0 < i0, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < i1 ∧ 0 < j → (i + i1 * j) |%| j = i |%| j ⊦ (i + i0 * j - j) |%| j = i |%| j, i0 * j = (i0 - 1) * j + 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%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs