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Proof step 21 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
# int2bits(j) = n * 8, 0 ≤ i, i→n(i) + 1 < n, i→n(i) < n, 0 < j, 8 * i < abs(8 * (i + 1) - 1), 8 * i + 1 < 8 * (i + 1), n * 8 ≤ 1 + (i→n(8 * (i + 1)) - 1) ⊦ n = i→n(i) + 1

Rule argument

Additional information

Used simplifier rules: ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j;% ⊦ ~1 < i ↔ 0 ≤ i;% ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);% ⊦ 1 ≤ i ↔ 0 < i;