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Proof step 41 from proof int2bytes-len-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
j < 2 ^ (n→i(m) * 8 + n→i(n) - 1), 2 ^ (n→i(m) * 8 + n→i(n) - 2) ≤ j, n < 8, 0 < j, j < 2 ^ (8 * (i + 1) - 1), 0 ≤ i, 2 ^ (8 * i) ≤ j, n→i(m) * 8 + n→i(n) ≤ 8 * (i + 1), 0 ≤ j, 8 * i + 1 < n→i(m) * 8 + n→i(n), 1 < n→i(m) * 8 + n→i(n), 0 < n→i(m) * 8 + n→i(n), 8 * i + 1 < 8 * (i + 1), i ≤ n→i(m) ⊦ m * 8 + n = 0, m = i→n(i), n = 0, n = 1

Rule argument

Additional information

Used simplifier rules: ⊦ 0 ≤ n→i(n);% ⊦ j ≤ j0 ∧ j0 ≤ j1 → j ≤ j1; (forward)% ⊦ i * j = j * i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;