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Proof step 11 from proof bytes2int-len-one-01 (previous step)

bytes2int(bytes0) = 0, bytes2int(bytes0) * 256 < 128, 0 ≤ band(b2i(by), 255), bytes2int(bytes0) * 256 + band(b2i(by), 255) < 128, 0 ≤ bytes2int(bytes0) * 256 + band(b2i(by), 255) ⊦ bytes0.first = i2b(0), bytes0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
bytes2int(bytes0) = 0, 0 ≤ band(b2i(by), 255), band(b2i(by), 255) < 128 ⊦ bytes0 = [], bytes0.first = i2b(0)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i + j = j + i;% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ band(i, j) = band(j, i);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);