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Proof step 28 from proof int2bytes-len-le-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
int2bytes(i).first = i2b(~1), 2 ^ (8 * n→i(# int2bytes(i) - 1) - 1) + i < 0, 0 ≤ 2 ^ (8 * n→i(# int2bytes(i) - 1)) + i, 2 ≤ # int2bytes(i), 0 ≤ b2i(int2bytes(i).rest.first) ⊦ int2bytes(i).rest = [], # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n) - 1) + i < 0

Rule argument

Additional information

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