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Proof step 10 from proof abs-tree (previous step)

ad.size = # ad.entries, (((ino => ad * abs(sdt, ad.entries)) * D0))(dirs), ((abs(tree1, ino) * D1))(dirs), ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files ⊦ tree1.dir? ∧ ad.meta = tree1.meta

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
ad = mkdir(m, md, m0, m1, entries), m = # entries, ((abs(tree1, ino) * D1))(dirs), (((ino => ad * abs(sdt, entries)) * D0))(dirs), ∀ ino. ⌜ino⌝ ∈ tree1.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino0. ⌜ino0⌝ ∈ sdt.fids → ino0 ∈ files ⊦ tree1.dir? ∧ md = tree1.meta

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, m, md, m0, 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, m, md, m0, entries