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Proof step 29 from proof int2bytes-first-zerobyte-01 (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: ⊦ m * n = n * m;% ⊦ 0 + m = m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;% ⊦ m + n = m + n0 ↔ n = n0;