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Proof step 1 from proof int2bits-rec-neg-03 (conclusion)

⊦ j % 2 ≠ 0 ∧ j < ~1 → int2bits(j) = int2bits(j / 2 - 1) + true '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
j < ~1, j < 0 ⊦ int2bits(j) = int2bits(j / 2 - 1) + true ', j % 2 = 0

Rule argument

Additional information

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