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Proof step 23 from proof mod-mult-add (previous step)

0 < j, (i + (i0 + 1) * j - j) |%| j = i |%| j, 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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i + (i0 + 1) * j, j, i2] flags:

Additional information

Used simplifier rules: eq-succ: ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i + (i0 + 1) * j, j, i2