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Proof step 14 from proof byte-mult-256 (previous step)

j = i0 * 256 + j0 ∧ j0 ≤ 0 ∧ abs(j0) < 256 ∧ i0 ≤ 0 ∧ 0 < 256, j0 < 0, j < 0, 0 < i ⊦ i = 0, i2b(i * 256 + j) = i2b(j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
i0 * 256 + j0 < 0, j0 < 0, i0 < i, i0 ≤ 0, ~256 < j0, 0 < i ⊦ i2b(256 * (i0 + i) + j0) = i2b(i0 * 256 + j0), i = 0

Rule argument

Additional information

Used simplifier rules: i2b-noteq: ⊦ i2b(i) ≠ i2b(j) → i ≠ j;% ⊦ i = i + j ↔ j = 0;% ⊦ 0 < i → (i * j = i * j0 ↔ j = j0);% ⊦ i + j = i + i0 ↔ j = i0;% ⊦ j ≤ i ∧ i ≤ 0 → j ≤ 0; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ ~ i < j ↔ 0 < j + i;% ⊦ i < 0 → abs(i) = ~ i;% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j + i * j0 = i * (j + j0);