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Proof step 22 from proof fids-in (previous step)

str0 ∈ sdt0, fid0 ∈ sdt0[str0].fids, fid ∈ sdt0.fids, str0 ∈ sdt0[str, tree], ∀ str0. str0 ∈ sdt0[str, tree] → (∀ fid. fid ∈ sdt0[str, tree][str0].fids → (∃ p. p ∈ sdt0[str, tree][str0] ∧ sdt0[str, tree][str0][p] = fnode(fid))), ∀ fid1. fid1 ∈ sdt0[str0].fids → (∃ p. p ∈ sdt0[str0] ∧ sdt0[str0][p] = fnode(fid1)), ∀ sdt1, fid, md. # sdt1 < # sdt0[str, tree] → fid ∈ sdt1.fids ∧ (∀ str. str ∈ sdt1 → (∀ fid. fid ∈ sdt1[str].fids → (∃ p. p ∈ sdt1[str] ∧ sdt1[str][p] = fnode(fid)))) → (∃ p. p ∈ dnode(md, sdt1) ∧ dnode(md, sdt1)[p] = fnode(fid)) ⊦ str ∈ sdt0, ∃ p. p ∈ sdt0[str0] ∧ sdt0[str0][p] = fnode(fid0), ∃ p. p ∈ dnode(md, sdt0[str, tree]) ∧ dnode(md, sdt0[str, tree])[p] = fnode(fid)

Proof step 22 from proof fids-in (previous step)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

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