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Proof step 5 from proof int2bytes-len-le-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ ¬ m < n ↔ n ≤ m;% ⊦ ¬ m ≤ n ↔ n < m;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward)% ⊦ i ≤ j ∧ j < j0 → i < j0; (forward)% ⊦ m < n → n ≠ 0; (forward)% ⊦ m < n → n ≠ 0;% ⊦ 0 < 2 ^ i ↔ 0 ≤ i;% ⊦ i ≤ j - i0 ↔ i + i0 ≤ j;% ⊦ 1 ≤ i ↔ 0 < i;% ⊦ 0 < i → (0 < i * j ↔ 0 < j);%0 ≤ i ⊦ (* i) < n→i(n) ↔ i→n(i) < n;% ⊦ 0 < n ↔ n ≠ 0;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ i * j = j * i;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1);% ⊦ i < j - i0 ↔ i + i0 < j;%0 ≤ j ⊦ 2 ^ j < 2 ^ j0 ↔ j < j0;% ⊦ i - j < i0 ↔ i < j + i0;% ⊦ i + (j - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0);% ⊦ i < j + 1 ↔ i ≤ j;% ⊦ n→i(m) ≤ n→i(n) ↔ m ≤ n;% ⊦ m < 1 ↔ m = 0;%n0 ≤ n ⊦ m ≤ n - n0 ↔ m + n0 ≤ n;% ⊦ m + m0 ≤ m ↔ m0 = 0;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ m + n = n + m;