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Proof step 28 from proof bytes2int-len-one-01 (previous step)

bytes0 = by0 + bytes, 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 = [], by0 = i2b(0), bytes2int(bytes0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
0 ≤ band(b2i(by), 255) + 256 * bytes2int(by0 ' + bytes), bytes2int(by0 ' + bytes) < 0, band(b2i(by), 255) + 256 * bytes2int(by0 ' + bytes) < 128, 0 ≤ band(b2i(by), 255) ⊦ bytes2int(by0 ' + bytes) = 0, by0 = i2b(0)

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ band(i, j) = band(j, i);% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' + bytes ≠ [];% ⊦ by + bytes = by ' + bytes;