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Proof step 8 from proof subset-append-02 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 9:
∀ a. a ∈ x ∨ a ∈ y → a ∈ z ⊦ (∀ a. a ∈ x → a ∈ z) ∧ (∀ a. a ∈ y → a ∈ z)
Proof step 15:
(∀ a. a ∈ x → a ∈ z) ∧ (∀ a. a ∈ y → a ∈ z) ⊦ ∀ a. a ∈ x ∨ a ∈ y → a ∈ z

Rule argument

pos: right 1

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