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Proof step 9 from proof px (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) ⊦ (∀ q. q ⊑ p → px(user, meta(q, tree, abs-files(files)))) ↔ (∀ q. q ⊑ p → px(user, meta(q, ino, dirs, files)))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information