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Proof step 3 from proof Del-empty (previous step)

⊦ ∀ str0. (str0 ∈ ∅ -- str ↔ str0 ∈ ∅) ∧ (∅ -- str)[str0] = (∅)[str0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∅ -- str)[str0] = (∅)[str0], str0 ∈ ∅ -- str

Rule argument

Additional information

Used simplifier rules: ⊦ str ∈ sdt -- str0 ↔ str ≠ str0 ∧ str ∈ sdt; (ws)% ⊦ ¬ str ∈ ∅; (s)