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Proof step 24 from proof int2bytes-last-01 (previous step)

~256 ≤ i ∨ i % 256 = 0, ~256 ≤ i ∨ i % 256 ≠ 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ (int2bytes(i / 256) + int2bytes(i % 256).last ').last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
~256 ≤ i ∨ i % 256 = 0, ~256 ≤ i ∨ i % 256 ≠ 0, i ≤ ~256 ∨ ~128 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i ⊦ int2bytes(i % 256).last = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').last = by;