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Proof step 28 from proof int2bytes-last-01 (previous step)

~256 = i, ~256 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, ~256 = i ∨ ~128 ≤ i ⊦ int2bytes(i % 256).last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
⊦ i2b(0) = i2b(~256)

Rule argument

Additional information

Used simplifier rules: int2bytes-byte: ~128 ≤ i, i < 128 ⊦ int2bytes((* i)) = i2b(i) ';% ⊦ by '.last = by;