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Proof step 67 from proof bytes2int-int2bytes-len-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ i2b(i - j) = i2b(i - b2i(i2b(j))) subst: [j, i] → [256, band(b2i(by), 255)] rotate: F spec: byte inst: name: byte-law-sub-one-01

Additional information

Used simplifier rules: ⊦ 0 ≤ i ∧ i % 256 ≤ 127 → b2i(i2b(i)) = i % 256;% ⊦ i - 0 = i;% ⊦ i2b(band(255, b2i(by))) = by;% ⊦ band(i, band(i0, j)) = band(band(i, i0), j);% ⊦ band(i, j) = band(j, i); Used terms: 256, band(b2i(by), 255)