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Proof step 11 from proof del-eq (previous step)

a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 = a, a2 ∈ s ↔ a2 ∈ s0, a2 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 ∈ s ↔ a2 ∈ s0, a2 = a

Rule argument

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