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Proof step 14 from proof int2bytes-len-le-neg (previous step)

2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, i < 0, 1 ≤ # int2bytes(i) ⊦ b2i(int2bytes(i).first) = ~1, n ≤ # int2bytes(i) ↔ ¬ 0 ≤ i + 2 ^ (8 * (n→i(n) - 1)), int2bytes(i).first = i2b(~1)

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 ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ i < 0 ∧ 0 ≤ j → i < j;% ⊦ i + j < i + i0 ↔ j < i0;% ⊦ 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 - i) = j;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ i + j = j + i;% ⊦ n→i(m) < n→i(n) ↔ m < n;% ⊦ m ≤ n → (n < m ↔ false);