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Proof step 120 from proof bytes2int-int2bytes-len-eq (previous step)

bytes1 = by0 + bytes, bytes1 = by0 ', ~256 < bytes2int(bytes1) * 256 + band(b2i(by), 255), bytes2int(bytes1) * 256 + band(b2i(by), 255) < 0, bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~128, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ bytes1 = [], bytes1 = i2b(~1) '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 121:
256 * b2i(by0) + band(b2i(by), 255) < ~128, ~256 < 256 * b2i(by0) + band(b2i(by), 255), 256 * b2i(by0) + band(b2i(by), 255) < 0, ∀ bytes0. # bytes0 < #((by0 + []) + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ i2b(~1) = by0

Rule argument

Additional information

Used simplifier rules: bytes2int-one: ⊦ bytes2int(by ') = b2i(by);% ⊦ bytes + [] = bytes;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)% ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ band(i, j) = band(j, i);% ⊦ by ' + bytes ≠ [];% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ by ' = by0 ' + bytes ↔ by = by0 ∧ bytes = [];% ⊦ by + bytes = by ' + bytes;