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Proof step 58 from proof abs-tree (previous step)

((ino => mkdir(# entries, md, m0, m1, entries) * (abs(sdt0 -- str, entries -- str) * abs(sdt0[str], entries[str])) * D1))(dirs), ((((abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * ino => mkdir(# entries, md, m0, m1, entries)) * D0))(dirs), ⌜ino1⌝ ∈ sdt[str].fids, str ∈ entries, str ∈ sdt, str ∈ sdt0, ∀ ino0. ⌜ino0⌝ ∈ sdt0.fids → ino0 ∈ 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 ⊦ ino1 ∈ files, (p0 ∈ sdt[str] ↔ p0 ∈ sdt0[str]) ∧ ( p0 ∈ sdt[str] → (sdt[str][p0].dir? ↔ sdt0[str][p0].dir?) ∧ (sdt[str][p0].dir? ⊃ sdt[str][p0].meta = sdt0[str][p0].meta;sdt[str][p0].fid = sdt0[str][p0].fid))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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