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Proof step 17 from proof again-put (previous step)

str ∈ sdt, p0 ∈ sdt[str], ∀ tree. p0 ∈ tree → tree[p0, tree[p0]] = tree ⊦ dnode(md, sdt[str, sdt[str][p0, sdt[str][p0]]]) = dnode(md, sdt), p0 = ε

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: sdt[str]

Additional information

Used simplifier rules: ⊦ sdt[str] = tree ∧ str ∈ sdt → sdt[str, tree] = sdt; (s) Used terms: sdt[str]