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Proof step 2 from proof subdef (previous step)

⊦ s1 ⊆ s2, s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s1 → a ∈ s2 ⊦ s1 ⊆ s2

Rule argument

Additional information