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Proof step 3 from proof del-eq (previous step)

a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0 ⊦ (∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)) ↔ (∀ a. a ∈ s ↔ a ∈ s0)

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 4:
a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ ∀ a. a ∈ s ↔ a ∈ s0
Proof step 10:
a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0)

Rule argument

pos: right 1

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