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Proof step 42 from proof int2bytes-last-01 (previous step)

0 ≤ i % 256 ∧ i % 256 ≤ 127, 256 ≤ i, 0 < i, 0 ≤ i ⊦ ~256 = i, i2b(i % 256) '.last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
i % 256 ≤ 127, 256 ≤ i, 0 < i, 0 ≤ i ⊦ ~256 = i, i2b(i % 256) = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < j ∧ 0 ≤ i → 0 ≤ i % j;% ⊦ by '.last = by;