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Proof step 25 from proof byte-law-mult-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
j0 < 256, 0 ≤ i0, 0 ≤ j0 ⊦ i2b((i0 * 256 + j0) * j) = i2b(b2i(i2b(j0)) * j)

Rule argument

Additional information

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