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Proof step 27 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < i ∧ 0 ≤ j → ((* j) < (* i) * i0 ↔ j / i < i0);% ⊦ i < j - i0 ↔ i + i0 < j;% ⊦ 1 < 2 ^ i ↔ 0 < i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ~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;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ int2bits(j) ≠ [];% ⊦ # bits = 0 ↔ bits = [];% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);