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Proof step 14 from proof int2bytes-first-zerobyte-01 (previous step)

0 ≤ i0 * 8 ∧ 0 ≤ j0, # int2bits(j) = i→n(i0 * 8) + i→n(j0), j0 < 8, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);