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Proof step 1 from proof size-one (conclusion)

⊦ # s = 1 → (∃ a. s = ∅ ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s = 1 ⊦ ∃ a. ∀ a0. a0 ∈ s ↔ a0 = a

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2);%in-04: ⊦ a ∈ ∅ ++ b ↔ a = b;