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Proof step 17 from proof int2bytes-rec-case (previous step)

i ≠ ~256 ∧ i % 256 = 0, i < 0, i ≤ ~256 ⊦ int2bytes(i) = int2bytes(i / 256) + int2bytes(i % 256).last '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i % 256 = 0, i < ~256, i < 0 ⊦ i = ~256, int2bytes(i) = int2bytes(i / 256) + i2b(0) '

Rule argument

Additional information

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