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Proof step 6 from proof subset-rec (previous step)

⊦ (∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y) ↔ a ∈ y ∧ (∀ a. a ∈ x → a ∈ y)

User applied rule case distinction

And obtained the following new goals

Proof step 7:
∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y ⊦ a ∈ y ∧ (∀ a. a ∈ x → a ∈ y)
Proof step 11:
a ∈ y ∧ (∀ a. a ∈ x → a ∈ y) ⊦ ∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y

Rule argument

pos: right 1

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