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Proof step 2 from proof fids-in (previous step)

fid ∈ tree.fids ⊦ ∃ p. p ∈ tree ∧ tree[p] = fnode(fid)

User applied rule induction

And obtained the following new goals

Proof step 3:
∀ fid. Ptree(fnode(fid)), ∀ md, sdt. (∀ str. str ∈ sdt → Ptree(sdt[str])) → Ptree(dnode(md, sdt)) ⊦ Ptree(tree)
Proof step 4:
⊦ (∀ fid. ∀ fid0, tree. fnode(fid) = tree ∧ fid0 ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid0))) ∧ (∀ md, sdt. (∀ str. str ∈ sdt → (∀ fid, tree. sdt[str] = tree ∧ fid ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid)))) → (∀ fid, tree. dnode(md, sdt) = tree ∧ fid ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid))))

Rule argument

spec: directory-tree name: tree-induction seq: ∀ fid. Ptree(fnode(fid)), ∀ md, sdt. (∀ str. str ∈ sdt → Ptree(sdt[str])) → Ptree(dnode(md, sdt)) ⊦ Ptree(tree) subst: [Ptree, tree] → [λ tree0. ∀ fid, tree. tree0 = tree ∧ fid ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid)), tree]

Additional information

Used terms: λ tree0. ∀ fid, tree. tree0 = tree ∧ fid ∈ tree.fids → (∃ p. p ∈ tree ∧ tree[p] = fnode(fid)), tree