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Proof step 31 from proof int2bytes-len-le-neg-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
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 = []

Rule argument

Additional information

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