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Proof step 24 from proof int2bytes-len-pos-01 (previous step)

i + 1 ≤ n→i(m), 0 < n→i(m) * 8 + n→i(n), 8 * (i + 1) < n→i(m) * 8 + n→i(n), 0 ≤ j, n→i(m) * 8 + n→i(n) < 2 + 8 * (i + 1), 2 ^ (8 * (i + 1) - 1) ≤ j, j < 2 ^ (8 * (i + 1)), 0 ≤ i, n < 8, 2 ^ (n→i(m) * 8 + n→i(n) - 2) ≤ j, j < 2 ^ (n→i(m) * 8 + n→i(n) - 1) ⊦ m + 1 = i→n(i + 2), n = 0, m * 8 + n = 0, m = 0 ∧ n = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
i < n→i(m), j < 2 ^ (n→i(m) * 8 + n→i(n) - 1), 2 ^ (n→i(m) * 8 + n→i(n) - 2) ≤ j, n < 8, 0 ≤ i, j < 2 ^ (8 * (i + 1)), 2 ^ (8 * (i + 1) - 1) ≤ j, n→i(m) * 8 + n→i(n) < 2 + 8 * (i + 1), 0 ≤ j, 8 * (i + 1) < n→i(m) * 8 + n→i(n), 0 < n→i(m) * 8 + n→i(n) ⊦ m = 0, m * 8 + n = 0, n = 0, m + 1 = i→n(i + 2)

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 0 + m = m;% ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + 1 ≤ j ↔ i < j;