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Proof step 14 from proof mod-pot-shift-pos (previous step)

0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → (i % j) ^ i1 % j = i ^ i1 % j ∨ i1 = 0 ⊦ ((i % j) ^ i2 % j * (i % j) % j) % j = (i ^ i2 * i) % j, ~1 = i2, 0 = j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 < j, 0 ≤ i2, ∀ i1, j, i. i→n(i1) < i→n(i2 + 1) → 0 < j ∧ 0 < i1 → (i % j) ^ i1 % j = i ^ i1 % j ∨ i1 = 0 ⊦ ((i % j) ^ i2 % j * i % j) % j = (i ^ i2 * i) % j, 0 = j, ~1 = i2

Rule argument

Additional information

Used simplifier rules: j ≠ 0 ⊦ (i % j) % j = i % j; (s)