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Proof step 42 from proof int2bytes-len-neg (previous step)

j < ~1 ∧ 0 ≤ 8 * (i + 1) - 1, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, # int2bits(j) ≤ i→n(8 * (i + 1) - 1) + 1, j < 0, 1 + i→n(i) * 8 < # int2bits(j) ⊦ j + 2 ^ (8 * i) = 0, # int2bits(j) / 8 = i→n(i), # int2bits(j) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 43:
i→n(i) * 8 < i→n(8 * (i + 1) - 1), j < ~1, 2 ^ (8 * i) ≤ 2 ^ (8 * (i + 1) - 1), 0 ≤ i, # int2bits(j) ≤ i→n(8 * (i + 1) - 1) + 1, j < 0, 1 + i→n(i) * 8 < # int2bits(j) ⊦ j + 2 ^ (8 * i) = 0, # int2bits(j) / 8 = i→n(i), # int2bits(j) % 8 = 0, 8 * (i + 1) = 1

Rule argument

Additional information

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