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Proof step 37 from proof abs-acyclic (previous step)

entries[str] = ROOT_INO, tree[p] = dnode(md0, sdt), (( ino => mkdir(# entries, md0, # links(ino, dirs), m2, entries) * (abs(sdt -- str, entries -- str) * abs(sdt[str], ROOT_INO)) * abs|(tree, ROOT_INO, p, ino)) )(dirs), abs(tree, ROOT_INO)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ROOT_INO, ino, dirs, files), str ∈ entries, str ∈ sdt ⊦ sdt[str].dir?, ispath(p + str ', ROOT_INO, ROOT_INO, dirs, files), str ∈ dirs[ino].entries ∧ dirs[ino].entries[str] = ROOT_INO

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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[ino], m, md0, m1, m2, entries0]

Additional information

Used simplifier rules: sep-ass: ⊦ (D * D0) * D1 = D * D0 * D1; from specification heap-sep[dir-heap]%sep-com: ⊦ D * D0 = D0 * D; from specification heap-sep[dir-heap]% ⊦ ((ino => ad * D))(dirs) → (dirs[ino] = ad0 ↔ ad = ad0); (s)%⊦ mkdir(m, md, m0, m1, entries) = mkdir(m2, md0, m3, m4, entries0) ↔ m = m2 ∧ md = md0 ∧ m0 = m3 ∧ m1 = m4 ∧ entries = entries0; (ws)%⊦ ( ((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[ino], m, md0, m1, m2, entries0