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Proof step 7 from proof dir-empty (previous step)

dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), tree[p + str '].dir?, tree[p].dir?, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), m ∈ dirs, str ∈ entries, entries[str] ∈ dirs ⊦ dirs[entries[str]].entries = ∅ ↔ tree[p + str '].entries = ∅

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
dirs[m] = mkdir(# entries, md, # links(m, dirs), n1, entries), dirs[entries[str]] = mkdir(m0, md0, m1, m2, entries0), tree[p].dir?, tree[p + str '].dir?, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), px-parent(user, p, ino, dirs, files), entries[str] ∈ dirs, str ∈ entries, m ∈ dirs ⊦ entries0 = ∅ ↔ tree[p + str '].entries = ∅

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[entries[str]], m0, md0, 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[entries[str]], m0, md0, m1, m2, entries0