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Proof step 17 from proof byte-mult-256 (previous step)

¬(i0 < 0 ∧ j0 < 0), 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
j0 < 0, ~256 < j0, 0 < i ⊦ i2b(256 * i + j0) = i2b(j0), i = 0

Rule argument

Additional information

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