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Proof step 9 from proof mod-pot-shift-pos (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 10:
i0 = i2 + 1, 0 < j, 0 < i0, ∀ i1, j, i. i→n(i1) < i→n(i0) → 0 < j ∧ 0 < i1 → i1 = 0 ∨ (i % j) ^ i1 % j = i ^ i1 % j ⊦ i0 = 0, ((i % j) ^ i2 * i % j) % j = i ^ i0 % j

Rule argument

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

Additional information

Used simplifier rules: ⊦ i0 = i - j ↔ i = i0 + j; (elim) Used terms: i0, 1, i2