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Proof step 5 from proof px-prefix (previous step)

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

User applied rule apply lemma

And obtained the following new goals

Proof step 6:
⊦ px(user, p, tree, fs) ∧ q ⊑ p → px(user, q, tree, fs)
Proof step 7:
⊦ 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))

Rule argument

spec: path-permission, inst: posix-permission name: px-prefix seq: ⊦ px(user, p, tree, fs) ∧ q ⊑ p → px(user, q, tree, fs) subst: [user, p, tree, fs, q] → [user, p, tree, abs-files(files), q]

Additional information

Used simplifier rules: 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)); Used terms: user, p, tree, abs-files(files), q