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Proof step 6 from proof abs-path-h (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 7:
dirs[ino] = mkdir(m, md, m0, m1, entries), tree = dnode(md0, sdt), ((abs(tree, ino) * D))(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), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ino ∈ dirs ∧ str ∈ entries ∧ ispath(p0, entries[str], dirs, files) ↔ str ' ∈ tree ∧ p0 ∈ tree[str ']
Proof step 16:
dirs[ino] = mkdir(m, md, m0, m1, entries), tree = fnode(fid), ((abs(tree, ino) * D))(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), ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ ino ∈ dirs ∧ str ∈ entries ∧ ispath(p0, entries[str], dirs, files) ↔ str ' ∈ tree ∧ p0 ∈ tree[str ']

Rule argument

var: tree

Additional information

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