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Proof step 195 from proof bytes2int-int2bytes-len-eq (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

seq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j subst: [j, i] → [i + 256, b2i(by)] rotate: F spec: byte inst: name: i2b-noteq

Additional information

Used simplifier rules: ⊦ i2b(b2i(by)) = by;% ⊦ i2b(i + 256) = i2b(i);% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i; Used terms: i + 256, b2i(by)