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Proof step 65 from proof abs-tree (previous step)

fid = ⌜m⌝ , (((m => mkdir(# entries, md, m0, m1, entries) * abs(sdt, entries)) * D0))(dirs), D1(dirs), str ∈ sdt, p0 ∈ sdt[str], str ∈ entries, ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ tree1, ino, D1, dirs, tree0, D0, files. ((abs(tree1, ino) * D1))(dirs) ∧ ((abs(tree0, ino) * D0))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files) ∧ (∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (p0 ∈ tree0 ↔ p0 ∈ tree1) ∧ (p0 ∈ tree0 → (tree0[p0].dir? ↔ tree1[p0].dir?) ∧ (tree0[p0].dir? ⊃ tree0[p0].meta = tree1[p0].meta;tree0[p0].fid = tree1[p0].fid)), ∀ ino0. ⌜ino0⌝ = fid → ino0 ∈ files ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
(((m => mkdir(# entries, md, m0, m1, entries) * abs(sdt, entries)) * D0))(dirs), D1(dirs), str ∈ entries, p0 ∈ sdt[str], str ∈ sdt, m ∈ files, ∀ tree1, ino, D1, dirs, tree0, D0, files. ((abs(tree1, ino) * D1))(dirs) ∧ ((abs(tree0, ino) * D0))(dirs) ∧ (∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files) ∧ (∀ ino. ⌜ino⌝ ∈ tree0.fids → ino ∈ files) ∧ (∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs) → (p0 ∈ tree0 ↔ p0 ∈ tree1) ∧ (p0 ∈ tree0 → (tree0[p0].dir? ↔ tree1[p0].dir?) ∧ (tree0[p0].dir? ⊃ tree0[p0].meta = tree1[p0].meta;tree0[p0].fid = tree1[p0].fid)), ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m⌝ = ⌜m0⌝ ↔ m = m0; (s)