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Proof step 4 from proof int2bits-falsein-02 (previous step)

¬ 0 ≤ j ⊦ false ∈ int2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j < 0 ⊦ false ∈ int2bits(j).rest ↔ (∀ i. 2 ^ i ≠ j + 1)

Rule argument

Additional information

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