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Proof step 4 from proof del-e-02 (previous step)

a0 ∈ s, a2 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a1 ∈ s ↔ a1 = a

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 5:
a2 ∈ s, a0 ∈ s, a1 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a1 = a
Proof step 6:
a1 = a, a2 ∈ s, a0 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a1 ∈ s

Rule argument

pos: right 1

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