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Proof step 16 from proof fids-strpath (previous step)

tree[p] = fnode(fid), p ∈ tree, ∀ sdt1, fid. # sdt1 < # sdt0[str, tree] → true → (fid ∈ sdt1.fids ↔ (∃ str, p. str ∈ sdt1 ∧ p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid))) ⊦ str ∈ sdt0, ∃ str0, p. str0 ∈ sdt0[str, tree] ∧ p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: str, p

Additional information

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