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Proof step 4 from proof size-one (previous step)

s = ∅ ∨ (∃ a. a ∈ s), # s = 1 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
# s = 1, (∀ a. ¬ a ∈ s) ∨ a ∈ s ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%In-empty: ⊦ ¬ a ∈ ∅;