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Proof step 19 from proof int2bytes-rec-neg-06 (previous step)

0 < i, j0 * 2 ^ (i * 8) + 2 ^ (i * 8 - 1) < 0 ⊦ (j0 * 2 ^ (i * 8)) % 2 ^ (i * 8) = 0 ∧ 1 - 2 ^ (i * 8) < 0 ∧ j0 * 2 ^ (i * 8) < 0 ∧ 0 < 2 ^ (i * 8), 2 ^ (i * 8) = 0, int2bytes(j0) = int2bytes(((1 - 2 ^ (i * 8)) + j0 * 2 ^ (i * 8)) / 2 ^ (i * 8))

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ i < j → (i ≠ j ↔ j < i);% ⊦ ¬ 2 ^ i < 0;% ⊦ i + j < j ↔ i < 0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;%j ≠ 0 ⊦ (i * j) % j = 0;% ⊦ i + (j - i0) = i + j - i0;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i + 0 = i;% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;