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Proof step 16 from proof bytes2int-len-one-02 (previous step)

256 * bytes2int((i2b(0) ' + by ') + bytes0) < 128, 0 ≤ band(255, b2i(by0)) + 256 * bytes2int((i2b(0) ' + by ') + bytes0), b2i(by) < 0, band(255, b2i(by0)) + 256 * bytes2int((i2b(0) ' + by ') + bytes0) < 128 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: bytes2int-le-zero: ⊦ bytes2int(i2b(0) ' + bytes) ≤ 0 ↔ bytes2int(i2b(0) ' + bytes) = 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i + j = j + i;% ⊦ band(i, j) = band(j, i);% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ i < 1 ↔ i ≤ 0;% ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);