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Proof step 1 from proof int2bits-len-neg-02 (conclusion)

⊦ ¬(∃ i. j = ~(2 ^ i)) ∧ j < 0 → # int2bits(j) = # int2bits(~ j)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j < 0 ⊦ # int2bits(j) = # int2bits(~ j), ∃ i. 2 ^ i + j = 0

Rule argument

Additional information

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