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Proof step 8 from proof int2bytes-len-eq-neg-01 (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];% ⊦ ¬ m < n → (¬ n < m ↔ n = m);% ⊦ ¬ n < n;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → (¬ ¬ j ≤ i ↔ false);% ⊦ ¬ m < n ↔ n ≤ m;% ⊦ m ≤ m;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i ≤ j → (¬ ¬ j < i ↔ false);% ⊦ i * j = j * i;% ⊦ # bytes = 0 ↔ bytes = [];