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Proof step 5 from proof mod-pred-neg (previous step)

abs(j0) < j, 0 < j, j0 ≤ 0, j0 < j, i0 ≤ 0, i0 < j, i0 * j + j0 < 0 ⊦ j0 = (i0 * j + j0 - 1) % j + 1, 1 = j0 + j

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
i0 * j + j0 = i1 + 1, abs(j0) < j, 0 < j, j0 ≤ 0, j0 < j, i0 ≤ 0, i0 < j, i0 * j + j0 < 0 ⊦ j0 = i1 % j + 1, 1 = j0 + j

Rule argument

spec: int-abs inst: name: elim-sub seq: ⊦ i0 = i - j ↔ i = i0 + j subst: [i, j, i0] → [i0 * j + j0, 1, i] flags:

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0 * j + j0, 1, i