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Proof step 51 from proof abs-nlinks-h (previous step)

abs(tree, ino0)(dirs), dirs[m].entries[str] ∈ files, ino0 ∈ dirs, m ∈ dirs, str ∈ dirs[m].entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ # links(dirs[m].entries[str], dirs) = 0, links(dirs[m].entries[str], dirs) = ∅, ⌜dirs[m].entries[str]⌝ ∈ tree.fids

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 52:
dirs[m] = mkdir(m0, md, m1, m2, entries), abs(tree, ino0)(dirs), str ∈ entries, m ∈ dirs, ino0 ∈ dirs, entries[str] ∈ files, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ # links(entries[str], dirs) = 0, links(entries[str], dirs) = ∅, ⌜entries[str]⌝ ∈ tree.fids

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] → [dirs[m], 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: dirs[m], m0, md, m1, entries