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Proof step 30 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
0 < j, 8 * i + 1 < 8 * (i + 1), j < 2 ^ (8 * (i + 1) - 1), 0 ≤ i, 2 ^ (8 * i) ≤ j ⊦ # int2bits(j) / 8 = i→n(i), # int2bits(j) % 8 = 0

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 0 = i;% ⊦ (* i) + i0 - (* j) = (i - j) + i0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 0 ≤ i → abs(i) = i;% ⊦ abs(i * j) = abs(i) * abs(j);% ⊦ n→i(i→n(i)) = abs(i);% ⊦ n→i(m + n) = n→i(m) + n→i(n);