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Proof step 192 from proof bytes2int-int2bytes-len-eq (previous step)

band(b2i(by), 255) = i + 256, bytes1 = int2bytes(bytes2int(bytes1)), 0 < 256 * (bytes2int(bytes1) + 1) + i, i < 0, 255 < 256 * (bytes2int(bytes1) + 1) + i, 0 ≤ bytes2int(bytes1), ~129 < i, ~256 ≤ i, 0 < bytes2int(bytes1), ∀ bytes0. # bytes0 < #(bytes1 + by ') → true → # bytes0 = # int2bytes(bytes2int(bytes0)) → bytes0 = int2bytes(bytes2int(bytes0)) ⊦ by = i2b(i), bytes1 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ b2i(by) < 0 → band(b2i(by), 255) = b2i(by) + 256 subst: [by] → [by] rotate: F spec: byte inst: name: byte-011x-neg

Additional information

Used simplifier rules: ⊦ i + j = i + i0 ↔ j = i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ b2i(by) = i → (i2b(i) ≠ by ↔ false); Used terms: by