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Proof step 4 from proof subset-empty-01 (previous step)

∀ a. ¬ a ∈ st ⊦ ∀ a. (a ∈ st ↔ a ∈ ∅) ∧ st[a] = (∅)[a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ st, ∀ a. ¬ a ∈ st ⊦

Rule argument

Additional information

Used simplifier rules: In-empty: ⊦ ¬ a ∈ ∅;%notmemlookup: ⊦ ¬ a ∈ st → st[a] = (∅)[a];