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Proof step 84 from proof abs-paths-eq (previous step)

((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3].dir?, tree[p3 + str ' + p0].dir?, ispath(p3, ino0, m, dirs, files), ispath(p0, dirs[m].entries[str], m, dirs, files), m ∈ dirs, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], ino0 ∈ dirs, p3 ∈ tree, str ∈ dirs[m].entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 85:
dirs[m] = mkdir(m0, md, m1, m2, entries), ((abs|(tree, ino0, p3, m) * abs(tree[p3], m)))(dirs), tree[p3 + str ' + p0].dir?, tree[p3].dir?, ispath(p0, entries[str], m, dirs, files), ispath(p3, ino0, m, dirs, files), str ∈ entries, p3 ∈ tree, ino0 ∈ dirs, p0 ∈ tree[p3][str '], str ' ∈ tree[p3], m ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦

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, m1, entries] → [dirs[m], m0, md, m1, m2, 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, m2, entries