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Proof step 2 from proof union-subset-02 (previous step)

∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, a0 ∈ s ∨ a0 ∈ s1 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2

User applied rule case distinction

And obtained the following new goals

Proof step 3:
a0 ∈ s, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2
Proof step 5:
a0 ∈ s1, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2

Rule argument

pos: left 3

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