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Proof step 32 from proof bytes2int-int2bytes-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ band(i, j) = band(j, i);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = m + n0 ↔ n = n0;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ # by ' = 0 +1;% ⊦ #(bytes + bytes0) = # bytes + # bytes0;