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Proof step 32 from proof int2bytes-len-le-neg-03 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information

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