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Proof step 43 from proof inv-mkdir (previous step)

# links(m, dirs) = dirs[m].nlink, dirs[m] = mkdir(dirs[m].size, md1, dirs[m].nlink, dirs[m].nsubdirs, entries), afs-cons(dirs, files), m ∈ dirs, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ str ∈ dirs[m].entries, str ∈ entries, ∃ ino. ¬ ino ∈ dirs ∧ ¬ ino ∈ files ∧ ino ≠ 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 44:
# links(m, dirs) = m1, dirs[m] = mkdir(m0, md1, m1, m2, entries), dirs[m] = mkdir(m0, md, m1, m2, entries0), afs-cons(dirs, files), m ∈ dirs, links(m, dirs) ≠ ∅ ∨ m = ROOT_INO ⊦ str ∈ entries, str ∈ entries0, ∃ ino. ¬ ino ∈ dirs ∧ ¬ ino ∈ files ∧ ino ≠ 0

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