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Proof step 2 from proof key-size (previous step)

∀ a. a ∈ ∅ ↔ a ∈ st0 ⊦ # ∅ = # st0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. ¬ a ∈ st0 ⊦ st0 = ∅

Rule argument

Additional information

Used simplifier rules: Size-empty: ⊦ # ∅ = 0;%empty: ⊦ # st = 0 ↔ st = ∅;%In-empty: ⊦ ¬ a ∈ ∅;