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

⊦ j % 2 = 0 ∧ j < 0 → int2bits(j) = int2bits(j / 2) + false '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j % 2 = 0, j < 0 ⊦ int2bits(j) = int2bits(j / 2) + false '

Rule argument

Additional information

Used simplifier rules: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward)