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Proof step 15 from proof int2bytes-len-le (previous step)

0 ≤ i, 2 ^ (8 * n→i(# int2bytes(i) - 1)) ≤ i ⊦ int2bytes(i).first = i2b(0), n ≤ # int2bytes(i) ↔ ¬ i < 2 ^ ((n→i(n) - 1) * 8)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 2

Additional information

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