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Proof step 46 from proof int2bytes-len-eq-pos (previous step)

int2bytes(i).first = i2b(0), 2 ^ (8 * (1 + n→i(# int2bytes(i).rest - 1)) - 1) ≤ i, 0 ≤ i, 2 ≤ # int2bytes(i), i < 2 ^ (8 * n→i(# int2bytes(i) - 1)) ⊦ int2bytes(i).rest = [], ¬ i < 2 ^ ((n→i(# int2bytes(i)) - 1) * 8 - 1) ∧ ¬ 2 ^ ((n→i(# int2bytes(i)) - 1) * 8) ≤ i

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ # bytes = 0 ↔ bytes = [];% ⊦ 0 < n ↔ n ≠ 0;%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j);%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ m ≤ m - 1 → m = 0; (forward)% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)