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Proof step 6 from proof count-entries (previous step)

∀ str. ¬ str ∈ sdt, ∀ entries0, sdt. # entries0 < # ∅ → (∀ str. str ∈ entries0 ↔ str ∈ sdt) → # entries0 = # sdt ⊦ ∀ str. (str ∈ sdt ↔ str ∈ ∅) ∧ sdt[str] = (∅)[str]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
str ∈ sdt, ∀ str. ¬ str ∈ sdt, ∀ entries0, sdt. # entries0 < # ∅ → (∀ str. str ∈ entries0 ↔ str ∈ sdt) → # entries0 = # sdt ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ str ∈ sdt → sdt[str] = (∅)[str]; (s)% ⊦ ¬ str ∈ ∅; (s)