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Proof step 94 from proof int2bytes-len-ls-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j);% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;%n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;