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Proof step 23 from proof abs-dir-h (previous step)

((ino => mkdir(# entries, md, m0, m1, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ sdt, str ∈ entries, ⌜ino1⌝ ∈ sdt[str].fids, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree, ino, D, dirs, ino0, files. ((abs(tree, ino) * D))(dirs) ∧ ispath(p0, ino, ino0, dirs, files) ∧ (∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (tree[p0].dir? ↔ ino0 ∈ dirs), ∀ ino1. ⌜ino1⌝ ∈ sdt.fids → ino1 ∈ files ⊦ sdt[str][p0].dir? ↔ ino0 ∈ dirs, ino1 ∈ files

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 8 terms: ino1

Additional information

Used simplifier rules: ⊦ fid ∈ sdt[str].fids ∧ str ∈ sdt → fid ∈ sdt.fids; (s) Used terms: ino1