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Proof step 183 from proof bytes2int-int2bytes-len-eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 184:
bytes1 = int2bytes(bytes2int(bytes1)), band(b2i(by), 255) < 256, 255 < bytes2int(bytes1) * 256 + band(b2i(by), 255), 0 < bytes2int(bytes1) * 256 + band(b2i(by), 255), 0 ≤ band(b2i(by), 255), 0 ≤ bytes2int(bytes1), ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by = int2bytes(band(b2i(by), 255)).last, bytes1 = []

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ bytes + by ' = bytes0 + by0 ' ↔ bytes = bytes0 ∧ by = by0;% ⊦ band(i, j) = band(j, i);