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Proof step 18 from proof inv-link (previous step)

fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ fid0 ∈ ofh.fids, root[path] = fnode(fid), root[parent(path')].dir?, root.dir?, fids-cons(root, fs, ofh), px(user, parent(path), root, fs), py(user, parent(path'), root, fs), parent(path') ∈ root, path ∈ root ⊦ fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ fid0 = fid ∨ fid0 ∈ ofh.fids, path' ∈ root

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)