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Proof step 20 from proof byte-law-add-one (previous step)

i = i0 * 256 + j0 ∧ 0 ≤ j0 ∧ j0 < 256 ∧ 0 ≤ i0 ∧ 0 < 256, j < 0, 0 ≤ i ⊦ j0 < 0, i2b(i + j) = i2b(b2i(i2b(i)) + j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
0 ≤ i0, 0 ≤ j0, j < 0, j0 < 256 ⊦ i2b(j0 + j) = i2b(b2i(i2b(j0)) + j)

Rule argument

Additional information

Used simplifier rules: byte-mult-256: ⊦ i2b(i * 256 + j) = i2b(j);% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0);% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;