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Proof step 25 from proof int2bytes-first-zerobyte-01 (previous step)

1 < 8 ∧ i→n(j0) < 8 → (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)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

Additional information

Used simplifier rules: ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;%0 ≤ i ⊦ i→n(i) < (* m) ↔ i < n→i(m);%j0 ≠ 0, j ≤ j0 ⊦ i < (* j) → i < (* j0);