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Proof step 52 from proof bytes2int-int2bytes-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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