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Proof step 37 from proof int2bytes-last-01 (previous step)

127 < i % 256 ∧ i % 256 ≤ 255, 0 < i % 256, 127 < i % 256, 0 ≤ i, 0 < i, 256 ≤ i ⊦ (i2b(0) ' + i2b(i % 256 - 256) ').last = i2b(i), ~256 = i

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 38:
i % 256 ≤ 255, 0 < i % 256, 127 < i % 256, 0 ≤ i, 0 < i, 256 ≤ i ⊦ ~256 = i, i2b(i % 256 - 256) = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').last = by;