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Proof step 48 from proof int2bytes-first-zerobyte-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
j < 2 ^ (8 * i0), 2 ^ (8 * i0 - 1) ≤ j, 0 ≤ j, 0 < i0 ⊦ i0 = 0, 8 * i→n(i0) = 1, ∃ i. ¬ i < 0 ∧ ¬ j < 2 ^ (8 * i - 1) ∧ ¬ 2 ^ (8 * i) ≤ j

Rule argument

Additional information

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