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Proof step 17 from proof inv-create (previous step)

fid0 ∈ fs ↔ fid0 ∈ root.fids ∨ fid0 ∈ ofh.fids, root.dir?, root[parent(path)].dir?, fids-cons(root, fs, ofh), py(user, parent(path), root, fs), parent(path) ∈ root ⊦ fid0 ∈ fs[fid, mkfile(md, mkbuf(0))] ↔ fid0 ∈ root.fids ∨ fid0 = fid ∨ fid0 ∈ ofh.fids, path ∈ root, fid ∈ fs

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ fid ∈ fs[fid0, f] ↔ fid = fid0 ∨ fid ∈ fs; (ws)% ⊦ fid ≠ fid0 → (fid ∈ fs[fid0, f] ↔ fid ∈ fs); (s)