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Proof step 2 from proof union-subset-04 (previous step)

∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s3 → a ∈ s4, a0 ∈ s ∨ a0 ∈ s1 ∨ a0 ∈ s3 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2, a0 ∈ s4

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, ∀ a. a ∈ s3 → a ∈ s4 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2, a0 ∈ s4
Proof step 5:
a0 ∈ s1, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s3 → a ∈ s4 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2, a0 ∈ s4
Proof step 7:
a0 ∈ s3, ∀ a. a ∈ s → a ∈ s0, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s3 → a ∈ s4 ⊦ a0 = a, a0 ∈ s0, a0 ∈ s2, a0 ∈ s4

Rule argument

pos: left 4

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