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Proof step 1 from proof subset-equal (conclusion)

⊦ s ⊆ s0 → (s0 ⊆ s ↔ s = s0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
∀ a. a ∈ s → a ∈ s0 ⊦ (∀ a. a ∈ s0 → a ∈ s) ↔ (∀ a. a ∈ s ↔ a ∈ s0)

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);