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Proof step 27 from proof abs-paths-eq (previous step)

tree[p3 + str ' + p0].dir?, tree[p3 + str0 ' + p1].dir?, abs(tree, ino0)(dirs), ispath(p3 + str0 ' + p1, ino0, ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p0, dirs[m].entries[str], ino1, dirs, files), ino1 ∈ dirs, p3 ∈ tree, str ' ∈ tree[p3], p0 ∈ tree[p3][str '], str0 ' ∈ tree[p3], p1 ∈ tree[p3][str0 '], ino0 ∈ dirs, m ∈ dirs, str ∈ dirs[m].entries, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files ⊦ str = str0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 28:
dirs[m] = mkdir(m0, md, m1, m2, entries), tree[p3 + str0 ' + p1].dir?, tree[p3 + str ' + p0].dir?, abs(tree, ino0)(dirs), ispath(p0, entries[str], ino1, dirs, files), ispath(p3, ino0, m, dirs, files), ispath(p3 + str0 ' + p1, ino0, ino1, dirs, files), str ∈ entries, m ∈ dirs, ino0 ∈ dirs, p1 ∈ tree[p3][str0 '], str0 ' ∈ tree[p3], p0 ∈ tree[p3][str '], str ' ∈ tree[p3], p3 ∈ tree, ino1 ∈ dirs, ∀ ino. ⌜ino⌝ ∈ tree.fids → ino ∈ files, ∀ ino. ¬ ino ∈ files ∨ ¬ ino ∈ dirs ⊦ str = str0

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