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Proof step 7 from proof int2bytes-len-le-neg-02 (previous step)

int2bytes(i) ≠ [], i < 0, bytes2int(int2bytes(i)) + 2 ^ (8 * n→i(# int2bytes(i).rest)) < 0 ⊦ int2bytes(i).first = i2b(~1), # int2bytes(i) ≤ n ↔ ¬ 2 ^ (8 * n→i(n)) + i < 0, b2i(int2bytes(i).first) = ~1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: int2bytes-empty: ⊦ int2bytes(i) ≠ [];%int2int: ⊦ bytes2int(int2bytes(i)) = i;% ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ i + j = j + i;% ⊦ i + j + i0 = (i + j) + i0;