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Proof step 2 from proof une-01 (previous step)

⊦ (∀ a. ¬ a ∈ s1 ∧ ¬ a ∈ s2) ↔ (∀ a. ¬ a ∈ s1) ∧ (∀ a. ¬ a ∈ s2)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. ¬ a ∈ s1 ∧ ¬ a ∈ s2 ⊦ (∀ a. ¬ a ∈ s1) ∧ (∀ a. ¬ a ∈ s2)
Proof step 9:
(∀ a. ¬ a ∈ s1) ∧ (∀ a. ¬ a ∈ s2) ⊦ ∀ a. ¬ a ∈ s1 ∧ ¬ a ∈ s2

Rule argument

pos: right 1

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