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Proof step 15 from proof int2bytes-len-le-neg-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
1 ≤ n, 2 ^ (8 * n→i(n)) + i < 0, 1 ≤ # int2bytes(i), i < 0, 2 ^ (8 * (n→i(# int2bytes(i)) - 1)) + i < 0, # int2bytes(i) ≤ n ⊦ int2bytes(i).first = i2b(~1), b2i(int2bytes(i).first) = ~1, n = 0

Rule argument

Additional information

Used simplifier rules: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward)% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;