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Proof step 33 from proof int2bytes-len-le-neg-03 (previous step)

¬ 2 ^ (8 * n→i(n) - 1) + i < 0, 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, int2bytes(i).rest = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
int2bytes(i).first = i2b(~1), n < # int2bytes(i), 2 ≤ # int2bytes(i), 0 ≤ b2i(int2bytes(i).rest.first), 1 ≤ # int2bytes(i), 2 ^ (8 * (n→i(# int2bytes(i)) - 1) - 1) + i < 0, 0 ≤ 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i, 0 ≤ 2 ^ (8 * n→i(n) - 1) + i, 2 ^ (8 * (n→i(# int2bytes(i)) - 1) - 1) < 2 ^ (8 * n→i(n) - 1) ⊦ int2bytes(i).rest = []

Rule argument

Additional information

Used simplifier rules: ⊦ i + j < i + i0 ↔ j < i0;% ⊦ i < j ∧ j ≤ j0 → i < j0; (forward)% ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;% ⊦ ¬ m ≤ n ↔ n < m;