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Proof step 48 from proof abs-tree (previous step)

((((abs(sdt -- str, entries -- str) * abs(sdt[str], entries[str])) * ino => mkdir(# entries, md, m0, m1, entries)) * D0))(dirs), ((ino => mkdir(# entries, md, m0, m1, entries) * abs(sdt0, entries) * D1))(dirs), str ∈ sdt0, str ∈ sdt, 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⌝ ∈ sdt0.fids → ino0 ∈ 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))

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: entries-recdel seq: ⊦ str ∈ sdt → abs(sdt, sdi) = ( ⌜str ∈ sdi⌝ ∧ abs(sdt -- str, sdi -- str) * abs(sdt[str], sdi[str])) subst: [str, sdi, sdt] → [str, entries, sdt0] flags: <all paths>

Additional information

Used terms: str, entries, sdt0