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Proof step 90 from proof inv-unlink (previous step)

# links(m, dirs) = dirs[m].nlink, dirs[m].size = 0, entries[str] = dirs[m].entries[str], dirs[m] = mkdir(0, md0, dirs[m].nlink, dirs[m].nsubdirs, entries), afs-cons(dirs, files), str ∈ dirs[m].entries, m ∈ dirs, str ∈ entries, entries[str] ∈ files, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 91:
# links(m, dirs) = m1, dirs[m] = mkdir(0, md0, m1, m2, entries), entries[str] = entries0[str], dirs[m] = mkdir(m0, md, m1, m2, entries0), m0 = 0, afs-cons(dirs, files), entries[str] ∈ files, str ∈ entries, m ∈ dirs, str ∈ entries0, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦

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, entries0]

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, entries0