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Proof step 19 from proof fids-strpath (previous step)

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

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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