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Proof step 2 from proof unsub-union-01 (previous step)

⊦ (∀ a. a ∈ s1 ∨ a ∈ s2 → a ∈ s) ↔ (∀ a. a ∈ s1 → a ∈ s) ∧ (∀ a. a ∈ s2 → a ∈ s)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s1 ∨ a ∈ s2 → a ∈ s ⊦ (∀ a. a ∈ s1 → a ∈ s) ∧ (∀ a. a ∈ s2 → a ∈ s)
Proof step 21:
(∀ a. a ∈ s1 → a ∈ s) ∧ (∀ a. a ∈ s2 → a ∈ s) ⊦ ∀ a. a ∈ s1 ∨ a ∈ s2 → a ∈ s

Rule argument

pos: right 1

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