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Proof step 7 from proof px-parent (previous step)

tree[p].dir? ↔ m ∈ dirs, abs(tree, ino)(dirs), afs-cons(dirs, files), fids-cons(tree, abs-files(files), ofh), ispath(p, ino, m, dirs, files), ino ∈ dirs ⊦ px(user, parent(p), tree, abs-files(files)) ↔ px(user, parent(p), ino, dirs, files), p = ε

User applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: path name: parent-elim seq: p ≠ ε ⊦ q = parent(p) ∧ str = p.last ↔ p = q + str ' subst: [p, q, str] → [p, q, str]

Additional information

Used simplifier rules: path-in: ⊦ ispath(p, ino, m, dirs, files) ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → p ∈ tree ∧ (tree[p].dir? ↔ m ∈ dirs);%path-in-eq: ⊦ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → (p ∈ tree ↔ ispath(p, ino, dirs, files));%ispath-target: ⊦ ispath(p, ino, dirs, files) ↔ (∃ m. ispath(p, ino, m, dirs, files));%px: ⊦ ispath(p, ino, m, dirs, files) ∧ abs(tree, ino)(dirs) ∧ fids-cons(tree, abs-files(files), ofh) ∧ afs-cons(dirs, files) → ( px(user, p, tree, abs-files(files)) ↔ px(user, p, ino, dirs, files));% ⊦ p + str ' ≠ ε; (s)%a: ⊦ (p + p0) + p1 = p + p0 + p1; from specification list[path]% ⊦ p + q ∈ tree ↔ p ∈ tree ∧ q ∈ tree[p]; (s)%p ≠ ε ⊦ q = parent(p) ∧ str = p.last ↔ p = q + str '; (elim) Used terms: p, q, str