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Proof step 15 from proof tree-induction (previous step)

∀ fid. Ptree(fnode(fid)), ∀ md, sdt. (∀ str. str ∈ sdt → Ptree(sdt[str])) → Ptree(dnode(md, sdt)), ∀ tree0, Ptree. #n tree0 < #n fnode(fid0) → (∀ md, sdt. (∀ str. str ∈ sdt → Ptree(sdt[str])) → Ptree(dnode(md, sdt))) ∧ (∀ fid. Ptree(fnode(fid))) → Ptree(tree0) ⊦ Ptree(fnode(fid0))

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: fid0

Additional information

Used terms: fid0