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Proof step 13 from proof abs-dir-h (previous step)

dirs[ino] = mkdir(m, md, m0, m1, entries), ((abs(tree, ino) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), str ∈ entries, ino ∈ dirs, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ 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) ⊦ tree[str ' + p0].dir? ↔ ino0 ∈ dirs

User applied rule constructor cut

And obtained the following new goals

Proof step 14:
dirs[ino] = mkdir(m, md, m0, m1, entries), tree = dnode(md0, sdt), ((abs(tree, ino) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), ino ∈ dirs, str ∈ entries, ∀ 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), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ tree[str ' + p0].dir? ↔ ino0 ∈ dirs
Proof step 24:
dirs[ino] = mkdir(m, md, m0, m1, entries), tree = fnode(fid), ((abs(tree, ino) * D))(dirs), ispath(p0, entries[str], ino0, dirs, files), ino ∈ dirs, str ∈ entries, ∀ 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), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ tree[str ' + p0].dir? ↔ ino0 ∈ dirs

Rule argument

var: tree

Additional information

Used simplifier rules: directory-tree, entries-tree generated by ∅, dnode, fnode, [ ];;