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Proof step 15 from proof inv-open (previous step)

fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids, root[path] = fnode(fid0), root.dir?, fids-cons(root, fs, ofh), path ∈ root, mode = MODE_W ∨ mode = MODE_RW → pw(user, path, root, fs), mode = MODE_R ∨ mode = MODE_RW → pr(user, path, root, fs) ⊦ fid ∈ fs ↔ fid ∈ root.fids ∨ fid = fid0 ∨ fid ∈ ofh.fids, n ∈ ofh

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ tree[p] = fnode(fid) ∧ p ∈ tree → fid ∈ tree.fids; (ss)