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Proof step 100 from proof fix-bits2bytes-01 (previous step)

n0 ≤ 8, # bits = m * 8 + n0, m < n, (m * 8 + n0) / 8 + 1 < n, n0 < 8 ⊦ (m * 8 + n0) % 8 = 0, n0 = 0, n * 8 = m * 8 + n0, bits = [], mklist(i2b(0), n - (m * 8 + n0) / 8 + 1) + bits2bytes(bits) = bits2bytes(mklist(false, (n - 1 - m) * 8) + mklist(false, 8 - n0) + bits), bits.first, (m * 8 + 8) % 8 = 0 ∧ (n - 1 - m) * 8 ≠ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 101:
(n - 1 - m) * 8 = 0, # bits = m * 8 + n0, 1 + m < n, m < n, n0 < 8 ⊦ bits = [], n * 8 = # bits, n0 = 0, mklist(i2b(0), n - 1 + m) + bits2bytes(bits) = bits2bytes(mklist(false, 8 - n0) + bits), bits.first

Rule argument

Additional information

Used simplifier rules: ⊦ (bits + bits0) + bits1 = bits + bits0 + bits1;% ⊦ [] + bits = bits;% ⊦ mklist(boolvar, 0) = [];% ⊦ m < n → m ≤ n;% ⊦ 0 + m = m;% ⊦ m < n → m / n = 0;%n ≠ 0 ⊦ (n0 + m * n) / n = n0 / n + m;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m < n → m % n = m;%n ≠ 0 ⊦ (m + n0 * n) % n = m % n;