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Proof step 58 from proof bytes2int-int2bytes-len-eq (previous step)

(bytes2int(bytes1) * 256) % 256 = 0 ∧ bytes2int(bytes1) * 256 + band(b2i(by), 255) < 0 ∧ band(b2i(by), 255) % 256 ≠ 0 ∧ bytes2int(bytes1) * 256 < 0 ∧ 0 ≤ band(b2i(by), 255) ∧ 0 < 256, # bytes1 = # int2bytes((bytes2int(bytes1) * 256) / 256 + band(b2i(by), 255) / 256 + 1 - 1), 0 ≤ band(b2i(by), 255), bytes2int(bytes1) < 0, bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~256, band(b2i(by), 255) < 256, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ band(b2i(by), 255) = 256, bytes2int(bytes1) * 256 + band(b2i(by), 255) = ~256, bytes1 = [], bytes1 + by ' = int2bytes((bytes2int(bytes1) * 256) / 256 + band(b2i(by), 255) / 256 + 1 - 1) + int2bytes(band(b2i(by), 255) - 256).last ', b2i(by) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ 0 < i → (i * j < 0 ↔ j < 0);% ⊦ i < 0 ∧ j0 < j ∧ 0 ≤ j → i * j + j0 < 0;% ⊦ band(i, j) = band(j, i);% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);%0 ≤ i, i < 128 ⊦ band(b2i(by), 255) = (* i) ↔ b2i(by) = i;%0 < j ⊦ 0 ≤ i ∧ i < j → i % (* j) = i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;%j ≠ 0 ⊦ (i * j) / j = i;%0 < j ⊦ 0 ≤ i ∧ i < j → i / (* j) = 0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;%i0 % j = 0, j ≠ 0 ⊦ (i * (* i0)) % (* j) = 0;