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Proof step 30 from proof int2bytes-first-zerobyte-01 (previous step)

0 * 8 + 1 = i→n(i0) * 8 + i→n(j0), 0 = i→n(i0) ∧ 1 = i→n(j0), 1 = i→n(j0) + 8 * i→n(i0), 0 ≤ i0, 0 ≤ j0, j0 < 8 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ 0 < i)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ m * n = n * m;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m);% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ i→n(i) = 0 ↔ i = 0;