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Proof step 141 from proof bytes2int-int2bytes-len-eq (previous step)

bytes1 = by0 + bytes, bytes2int(bytes1) * 256 + band(b2i(by), 255) = i + 256, bytes1 = by0 ', ~129 < i, i < 0, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 + by ' = i2b(0) ' + i2b(i) ', bytes1 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 142:
256 * b2i(by0) + band(b2i(by), 255) = i + 256, i < 0, ~129 < i, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by0 ' + by ' = i2b(0) ' + i2b(i) '

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ [] + bytes = bytes;% ⊦ bytes + [] = bytes;% ⊦ by ' + bytes ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ by + bytes = by ' + bytes;