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Proof step 10 from proof del-e-02 (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 11:
∀ a0. ¬ a0 ∈ s, a0 ∈ s, a1 ∈ s ⊦ a0 = a
Proof step 12:
∀ a0. a0 ∈ s ↔ a0 = a, a0 ∈ s, a1 ∈ s ⊦ a0 = a

Rule argument

pos: left 3

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