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Proof step 1 from proof int2bytes-len-ls-neg (conclusion)

⊦ i < 0 ∧ int2bytes(i).first ≠ i2b(~1) → (n < # int2bytes(i) ↔ i < ~(2 ^ (8 * n→i(n))))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < ~ j ↔ i + j < 0;