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Proof step 10 from proof subset-append-02 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 11:
∀ a. a ∈ x ∨ a ∈ y → a ∈ z ⊦ a ∈ x → a ∈ z
Proof step 13:
∀ a. a ∈ x ∨ a ∈ y → a ∈ z ⊦ a0 ∈ y → a0 ∈ z

Rule argument

pos: right 1

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