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Proof step 6 from proof size-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# s = 1, a ∈ s ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

Additional information

Used simplifier rules: size-one-02: ⊦ # s = 1 ∧ a ∈ s ∧ b ∈ s → a = b;%ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);