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Proof step 16 from proof subset-append-02 (previous step)

a ∈ x ∨ a ∈ y, ∀ a. a ∈ x → a ∈ z, ∀ a. a ∈ y → a ∈ z ⊦ a ∈ z

User applied rule case distinction

And obtained the following new goals

Proof step 17:
a ∈ x, ∀ a. a ∈ x → a ∈ z, ∀ a. a ∈ y → a ∈ z ⊦ a ∈ z
Proof step 18:
a ∈ y, ∀ a. a ∈ x → a ∈ z, ∀ a. a ∈ y → a ∈ z ⊦ a ∈ z

Rule argument

pos: left 1

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