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Proof step 3 from proof dom-isemp (previous step)

⊦ (∀ a. a ∈ dom st ↔ a ∈ ∅) ↔ st = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. ¬ a ∈ st ⊦ st = ∅

Rule argument

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st;% ⊦ ¬ a ∈ ∅; (s)% ⊦ ¬ a ∈ ∅; (s)