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Proof step 26 from proof bytes2int-len-one-01 (previous step)

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) ⊦ bytes2int(bytes0) = 0, bytes0.first = i2b(0), bytes0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
bytes2int(bytes0) < 0, 0 ≤ band(b2i(by), 255), 0 ≤ bytes2int(bytes0) * 256 + band(b2i(by), 255), bytes2int(bytes0) * 256 + band(b2i(by), 255) < 128 ⊦ bytes0 = [], bytes0.first = i2b(0), bytes2int(bytes0) = 0

Rule argument

Additional information

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