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Proof step 3 from proof fid-file (previous step)

abs(tree, ino)(dirs), fids-cons(tree, abs-files(files), ofh), ⌜n⌝ ∈ tree.fids ⊦ n ∈ files

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ fids-cons(root, fs, ofh) ↔ (∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids)
Proof step 5:
abs(tree, ino)(dirs), ⌜n⌝ ∈ tree.fids, ∀ fid. fid ∈ abs-files(files) ↔ fid ∈ tree.fids ∨ fid ∈ ofh.fids ⊦ n ∈ files

Rule argument

spec: posix-base name: fids-cons seq: ⊦ fids-cons(root, fs, ofh) ↔ (∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids) subst: [ofh, fs, root] → [ofh, abs-files(files), tree] flags: <all paths>

Additional information

Used terms: ofh, abs-files(files), tree