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Proof step 5 from proof band-mod256-pos (previous step)

0 ≤ i % 256 ∧ 0 ≤ 8 ⊦ 0 ≤ i → (i % 256) % 2 ^ 8 = i % 256

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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