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Proof step 67 from proof byte-mult-256 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 68:
i0 + i < 0, j0 < 256, 0 ≤ j0, 0 ≤ i0, i < 0 ⊦ i0 + i = 0, i = 0, i2b(256 * (i0 + i) + j0) = i2b(j0)

Rule argument

Additional information

Used simplifier rules: ⊦ i + j < j ↔ i < 0;% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;