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Proof step 16 from proof int2bytes-len-pos-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
0 ≤ i, j < 2 ^ (8 * (i + 1)), 2 ^ (8 * (i + 1) - 1) ≤ j ⊦ # int2bits(j) % 8 = 0, # int2bits(j) / 8 + 1 = i→n(i + 2)

Rule argument

Additional information

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