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Proof step 2 from proof subset-equal (previous step)

∀ a. a ∈ s → a ∈ s0 ⊦ (∀ a. a ∈ s0 → a ∈ s) ↔ (∀ a. a ∈ s ↔ a ∈ s0)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. a ∈ s0 → a ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ ∀ a. a ∈ s ↔ a ∈ s0
Proof step 7:
∀ a. a ∈ s ↔ a ∈ s0, ∀ a. a ∈ s → a ∈ s0 ⊦ ∀ a. a ∈ s0 → a ∈ s

Rule argument

pos: right 1

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