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Proof step 20 from proof int2bytes-first-zerobyte-01 (previous step)

¬(0 ≤ j ∧ 1 < i→n(j0) + 8 * i→n(i0)), # int2bits(j) = i→n(j0) + 8 * i→n(i0), 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

Proof step 21:
# int2bits(j) = i→n(j0) + 8 * i→n(i0), 0 ≤ i0, 0 ≤ j0, 0 ≤ j, j0 < 8 ⊦ j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j), (j0 ≠ 0 ∨ i0 ≠ 0) ∧ j ≠ 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ i → (# int2bits(i) = 1 ↔ i = 0);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;%0 ≠ m ⊦ (* m) * n = 0 ↔ n = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ ¬ i ≤ j ↔ j < i;