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Proof step 15 from proof abs-path-h (previous step)

((((abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * ino => mkdir(# entries, md, m0, m1, entries)) * D))(dirs), str ∈ sdt, str ∈ entries, ⌜ino0⌝ ∈ sdt[str].fids, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree, ino, D, dirs, files. ((abs(tree, ino) * D))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (ispath(p0, ino, dirs, files) ↔ p0 ∈ tree), ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files ⊦ ispath(p0, entries[str], dirs, files) ↔ p0 ∈ sdt[str], ino0 ∈ files

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: ino0

Additional information

Used simplifier rules: ⊦ fid ∈ sdt[str].fids ∧ str ∈ sdt → fid ∈ sdt.fids; (s) Used terms: ino0