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Proof step 4 from proof fids-exists (previous step)

⊦ (fnode(fid) = tree → (∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 = fid)) ∧ ((∀ str. str ∈ sdt → (∃ fids. ∀ fid. fid ∈ fids ↔ fid ∈ sdt[str].fids)) → dnode(md, sdt) = tree0 → (∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt.fids))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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