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Proof step 17 from proof int2bytes-last-01 (previous step)

¬(~128 ≤ i % 256 ∧ i % 256 < 128), i < 0, i ≤ ~256, i < ~256 ⊦ int2bytes(i % 256).last = i2b(i), i % 256 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
i % 256 < ~128, i < ~256, i ≤ ~256, i < 0 ⊦ i % 256 = 0, int2bytes(i % 256).last = i2b(i)

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j;%0 < j ⊦ i % (* j) < 0 ↔ i < 0 ∧ i % j ≠ 0;% ⊦ i < (* j) ∧ j < 0 → i < 0; (forward)%j0 ≠ 0, j < j0 ⊦ i < (* j) → i < (* j0);% ⊦ 0 < j ∧ i < 0 → (0 ≤ i % j ↔ i % j = 0);% ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;