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Proof step 35 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
# int2bits(j) = 1, 2 ^ (8 * i) ≤ j, 0 < j, 8 * i + 1 < 8 * (i + 1), j < 2 ^ (8 * (i + 1) - 1), 0 < i ⊦ i = 0

Rule argument

Additional information

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