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Proof step 76 from proof abs-tree (previous step)

ad.size = # ad.entries, (((m => ad * abs(sdt, ad.entries)) * D1))(dirs), D0(dirs), str ∈ ad.entries, m ∈ files, p0 ∈ sdt[str], str ∈ sdt, ∀ ino. ⌜ino⌝ ∈ sdt.fids → ino ∈ 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 ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 77:
ad = mkdir(m0, md, m1, m2, entries), m0 = # entries, (((m => ad * abs(sdt, entries)) * D1))(dirs), D0(dirs), str ∈ sdt, p0 ∈ sdt[str], m ∈ files, str ∈ entries, ∀ 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)), ∀ ino. ⌜ino⌝ ∈ sdt.fids → ino ∈ files ⊦

Rule argument

spec: abstract-dir name: elim seq: ⊦ ( ((m = ad.size ∧ md = ad.meta) ∧ m0 = ad.nlink) ∧ m1 = ad.nsubdirs) ∧ entries = ad.entries ↔ ad = mkdir(m, md, m0, m1, entries) subst: [ad, m, md, m0, entries] → [ad, m0, md, m1, entries]

Additional information

Used simplifier rules: ⊦ ( ((m = ad.size ∧ md = ad.meta) ∧ m0 = ad.nlink) ∧ m1 = ad.nsubdirs) ∧ entries = ad.entries ↔ ad = mkdir(m, md, m0, m1, entries); (elim) Used terms: ad, m0, md, m1, entries