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Proof step 63 from proof bytes2int-int2bytes-len-eq (previous step)

¬(~128 ≤ band(b2i(by), 255) - 256 ∧ band(b2i(by), 255) - 256 < 128), bytes1 = int2bytes(bytes2int(bytes1)), band(b2i(by), 255) < 256, bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~256, bytes2int(bytes1) < 0, 0 ≤ band(b2i(by), 255), ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by = int2bytes(band(b2i(by), 255) - 256).last, bytes1 = [], bytes2int(bytes1) * 256 + band(b2i(by), 255) = ~256, band(b2i(by), 255) = 256, b2i(by) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
bytes1 = int2bytes(bytes2int(bytes1)), bytes2int(bytes1) * 256 + band(b2i(by), 255) < ~256, bytes2int(bytes1) < 0, 0 ≤ band(b2i(by), 255), band(b2i(by), 255) < 128, ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by = int2bytes(band(b2i(by), 255) - 256).last, bytes1 = [], bytes2int(bytes1) * 256 + band(b2i(by), 255) = ~256, band(b2i(by), 255) = 256, b2i(by) = 0

Rule argument

Additional information

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