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Proof step 32 from proof int2bytes-first-zerobyte-01 (previous step)

0 ≤ j ∧ 1 < i→n(j0) + 8 * i→n(i0), 2 ^ (n→i(i→n(j0) + 8 * i→n(i0)) - 2) ≤ j ∧ j < 2 ^ (n→i(i→n(j0) + 8 * i→n(i0)) - 1), 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 33:
2 ^ (j0 + 8 * i0 - 2) < 2 ^ (j0 + 8 * i0 - 1), j < 2 ^ (j0 + 8 * i0 - 1), 2 ^ (j0 + 8 * i0 - 2) ≤ j, 0 ≤ i0, 0 ≤ j0, 0 ≤ j, j0 < 8, 0 < j0 + 8 * i0 ⊦ 8 * i→n(i0) = 1, i→n(j0) + 8 * i→n(i0) = 0, j0 = 1 ↔ (∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j), i0 = 0 ∧ j0 = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ m0 * m = 1 ↔ m0 = 1 ∧ m = 1;% ⊦ i + j = j + i;% ⊦ i * j = j * i;% ⊦ n→i(m * n) = n→i(m) * n→i(n);% ⊦ 0 ≤ i → n→i(i→n(i)) = i;% ⊦ n→i(m + n) = n→i(m) + n→i(n);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;%0 ≤ i ⊦ i→n(i) = (* m) ↔ i = n→i(m);% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ m * n = n * m;% ⊦ m0 * m * n = (m0 * m) * n;% ⊦ m0 * m = 0 ↔ m0 = 0 ∨ m = 0;% ⊦ i→n(i) = 0 ↔ i = 0;% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);