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Proof step 11 from proof size-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
# s = 1, ∀ a. ¬ a ∈ s ⊦ a ∈ s, ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

Additional information