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Proof step 67 from proof int2bytes-len-ls-01 (previous step)

1 ≤ n→i(n) * 8 ∧ 0 ≤ 1, # int2bits(i) = i→n(i0) * 8 + 1, # int2bits(i) / 8 = i→n(i0), i→n(i0) * 8 ≤ i→n(n→i(n) * 8) - i→n(1), 0 < i, i→n(i0) * 8 < # int2bits(i), i→n(i0) * 8 - 1 < i→n(i0) * 8, # int2bits(i) ≤ (i→n(n→i(n) * 8) - i→n(1)) + 1, n ≤ i→n(i0), 0 < i0 ⊦ n = 0, i0 = 0, i = 0, # int2bits(i) % 8 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

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