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Proof step 37 from proof abs-tree (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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