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Proof step 39 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 40:
# int2bits(i) / 8 = i→n(i0), i→n(8 * i0 - 1) + 1 < # int2bits(i), 0 < i, i < 2 ^ (8 * i0), 0 < i0 ⊦ # int2bits(i) % 8 = 0, i = 0, n→i(n) ≤ i0 ↔ ¬ i < 2 ^ (n→i(n) * 8 - 1), i0 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);