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Proof step 1 from proof int2bits-falsein-02 (conclusion)

⊦ false ∈ int2bits(j).rest ↔ ¬(∃ i. j = 2 ^ i - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ false ∈ int2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ i - j = i0 ↔ i = i0 + j;