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Proof step 14 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 - 1) + int2bytes(i % 256).last ').last = i2b(i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i < 0, i ≤ ~256, i < ~256 ⊦ int2bytes(i % 256).last = i2b(i), i % 256 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ j ≤ i ∧ i < 0 → j < 0; (forward)% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ i < j → ¬ j < i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1;% ⊦ (bytes + by ').last = by;