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Proof step 50 from proof abs-tree (previous step)

((ino => mkdir(# entries, md, m0, m1, entries) * (⌜str ∈ 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), 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 ⊦ (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

Proof step 51:
((((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 -- str, entries -- str) * abs(sdt0[str], entries[str])) * 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))

Rule argument

Additional information

Used simplifier rules: and-com: ⊦ (D ∧ D0) = (D0 ∧ D); from specification heap-sep[dir-heap]%and-ass: ⊦ ((D ∧ D0) ∧ D1) = (D ∧ D0 ∧ D1); from specification heap-sep[dir-heap]% ⊦ (true ∧ D) = D; (s)% ⊦ ⌜true⌝ = true; (s)