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Proof step 29 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)), 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)), bytes1 = [], bytes2int(bytes1) * 256 + band(b2i(by), 255) = ~256

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# bytes1 + 1 = # int2bytes(bytes2int(bytes1) * 256 + band(b2i(by), 255)), 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)), bytes1 = [], bytes2int(bytes1) * 256 + band(b2i(by), 255) = ~256, (bytes2int(bytes1) * 256 + band(b2i(by), 255)) % 256 = 0

Rule argument

Additional information

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