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Proof step 26 from proof int2bytes-last-01 (previous step)

~256 ≤ i, ~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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
~256 ≤ i, i < 128 ∨ 256 ≤ i, i < ~128 ∨ 128 ≤ i, ~256 = i ∨ ~128 ≤ i ⊦ int2bytes(i % 256).last = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i ≤ j → (j ≤ i ↔ i = j);%i0 ≠ 0, i0 ≤ i ⊦ (* i) ≤ j → (* i0) ≤ j;