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Proof step 3 from proof read-notin (previous step)

∀ str. sdt0[str] = (∅)[str] ∨ str ∈ sdt0 ⊦ sdt0[str0, tree][str] = (∅)[str], str ∈ sdt0[str0, tree]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ str. sdt0[str] = (∅)[str] ∨ str ∈ sdt0 ⊦ sdt0[str] = (∅)[str], str = str0, str ∈ sdt0

Rule argument

Additional information

Used simplifier rules: ⊦ str ≠ str0 → sdt[str0, tree][str] = sdt[str]; (s)% ⊦ str ∈ sdt[str0, tree] ↔ str = str0 ∨ str ∈ sdt; (ws)