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Proof step 3 from proof abs-has-rootpath-h (previous step)

((abs(tree, ino0) * {dirs0} ))(dirs), ino1 ∈ dirs ⊦ ino1 ∈ dirs0, ∃ p. ispath(p, ino0, ino1, dirs, files)

User applied rule induction

And obtained the following new goals

Proof step 4:
∀ fid. Ptree(fnode(fid)), ∀ md, sdt. (∀ str. str ∈ sdt → Ptree(sdt[str])) → Ptree(dnode(md, sdt)) ⊦ Ptree(tree)
Proof step 5:
⊦ (∀ fid. ∀ ino1, dirs, tree, ino0, dirs0, files. fnode(fid) = tree ∧ ino1 ∈ dirs ∧ ((abs(tree, ino0) * {dirs0} ))(dirs) → ino1 ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino1, dirs, files))) ∧ (∀ md, sdt. (∀ str. str ∈ sdt → (∀ ino1, dirs, tree, ino0, dirs0, files. sdt[str] = tree ∧ ino1 ∈ dirs ∧ ((abs(tree, ino0) * {dirs0} ))(dirs) → ino1 ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino1, dirs, files)))) → (∀ ino1, dirs, tree, ino0, dirs0, files. dnode(md, sdt) = tree ∧ ino1 ∈ dirs ∧ ((abs(tree, ino0) * {dirs0} ))(dirs) → ino1 ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino1, dirs, files))))

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. ∀ ino1, dirs, tree, ino0, dirs0, files. tree0 = tree ∧ ino1 ∈ dirs ∧ ((abs(tree, ino0) * {dirs0} ))(dirs) → ino1 ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino1, dirs, files)), tree]

Additional information

Used terms: λ tree0. ∀ ino1, dirs, tree, ino0, dirs0, files. tree0 = tree ∧ ino1 ∈ dirs ∧ ((abs(tree, ino0) * {dirs0} ))(dirs) → ino1 ∈ dirs0 ∨ (∃ p. ispath(p, ino0, ino1, dirs, files)), tree