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Proof step 61 from proof int2bytes-len-ls-01 (previous step)

0 ≤ i0, # int2bits(i) = i→n(i0) * 8 + 1, # int2bits(i) / 8 = i→n(i0), 0 < i, i→n(i0) * 8 < # int2bits(i), i→n(i0) * 8 - 1 < i→n(i0) * 8, i < 2 ^ (n→i(n) * 8 - 1), n ≤ i→n(i0), 0 < i0 ⊦ n = 0, i0 = 0, i = 0, # int2bits(i) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ i ≠ j → (i ≤ j ↔ i < j);