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Proof step 1 from proof size-one-03 (conclusion)

⊦ # s = 1 ↔ (∃ a. s = ∅ ++ a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
# s = 1 ⊦ ∃ a. s = ∅ ++ a

Rule argument

Additional information

Used simplifier rules: size-one-01: ⊦ #(∅ ++ a) = 1;