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Proof step 238 from proof inv-link (previous step)

dirs[ROOT_INO].nlink = 0, dirs[ino1] = mkdir(m1, md1, # links(ino1, dirs), m4, entries), files[entries0[str]] = mkfile(m, md, # links(entries0[str], dirs), content), dirs[ino0] = mkdir(m0, md0, # links(ino0, dirs), m3, entries0), p = p.head ', afs-cons(dirs, files), ispath(q, ROOT_INO, ino0, dirs, files), ispath(parent(p), ROOT_INO, ino1, dirs, files), of-cons(of, dirs, files), pw(user, md1), px(user, md0), px(user, md1), px-parent(user, q, ROOT_INO, dirs, files), px-parent(user, parent(p), ROOT_INO, dirs, files), str ∈ entries0, entries0[str] ∈ files, ino0 ∈ dirs, ino1 ∈ dirs, links(ino0, dirs) ≠ ∅ ∨ ino0 = ROOT_INO ⊦ ino1 = ROOT_INO, p = ε, entries0[str] ∈ dirs

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: list-data, inst: path name: elim-list-c seq: p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0 subst: [p, str, p0] → [p, str0, p0]

Additional information

Used simplifier rules: ⊦ str + p = str ' + p; (s)% ⊦ str ' = str0 ' + p ↔ str = str0 ∧ p = ε; (ws)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ str ' + p ≠ ε; (s)% ⊦ p + ε = p; (s)% ⊦ parent(str ') = ε; (s)% ⊦ ispath(ε, n, m, dirs, files) ↔ valid-ino(n, dirs, files) ∧ m = n; (s)% ⊦ valid-ino(ino, dirs, files) ↔ ino ∈ dirs ∨ ino ∈ files; (s)% ⊦ afs-cons(dirs, files) → ¬ ROOT_INO ∈ files; (s)%p ≠ ε ⊦ str = p.head ∧ p0 = p.tail ↔ p = str + p0; (elim) Used terms: p, str0, p0