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Proof step 90 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 91:
# int2bits(i) = i→n(i0) * 8 + 1, # int2bits(i) / 8 = i→n(i0), 0 < i, i0 < n→i(n), 2 ^ (n→i(n) * 8 - 1) ≤ i, 0 < i0, 2 ^ (i0 * 8 - 1) ≤ i ⊦ n = 1, n = 0, i0 = 0, i = 0, # int2bits(i) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ 0 < i → n→i(i→n(i)) = i;% ⊦ n→i(m * n) = n→i(m) * n→i(n);