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Proof step 20 from proof int2bytes-last-01 (previous step)

~256 < i % 256 ∧ i % 256 < ~128, i % 256 < ~128, i < ~256, i ≤ ~256, i < 0 ⊦ i % 256 = 0, (i2b(~1) ' + i2b(i % 256) ').last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
i < 0, i ≤ ~256, i < ~256, i % 256 < ~128, ~256 < i % 256 ⊦ i2b(i % 256) = i2b(i), i % 256 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;%j0 ≠ 0, j ≤ j0 ⊦ i ≤ (* j) → i ≤ (* j0);% ⊦ i + (* j) ≤ (* j0) ↔ i ≤ j0 - j;% ⊦ i ≤ ~ j ↔ i + j ≤ 0;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < 0 ∧ 0 ≤ j → i < j;%j ≠ 0 ⊦ i < i % j ↔ i < 0 ∧ (abs(j) = 1 ∨ abs(j) ≤ abs(i));% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').last = by;