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Proof step 61 from proof rename-fids (previous step)

fid ∈ fs ↔ fid ∈ root0.fids ∨ fid ∈ ofh.fids, root0[path'] = dnode(md0, sdt0), dnode(md, sdt) = root0[path], sdt0 = ∅, root0[path'].dir?, (root0 -- path)[parent(path')].dir?, root0[parent(path')].dir?, fids-cons(root0, fs, ofh), pw(user, path, root0, fs), py(user, parent(path), root0, fs), py(user, parent(path'), root0, fs), path' ∈ root0, path ∈ root0, parent(path') ∈ root0 ⊦ fid ∈ fs ↔ fid ∈ ((root0.fids \ sdt.fids) \ (path' ⊑ path ⊃ root0[path'] -- restn(# path', path);root0[path']).fids) ∪ sdt.fids ∨ fid ∈ ofh.fids, path' = ε, path ⊑ path'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
fid ∈ fs ↔ fid ∈ root0.fids ∨ fid ∈ ofh.fids, root0[path'] = dnode(md0, ∅), dnode(md, sdt) = root0[path], (root0 -- path)[parent(path')].dir?, root0[parent(path')].dir?, fids-cons(root0, fs, ofh), pw(user, path, root0, fs), py(user, parent(path), root0, fs), py(user, parent(path'), root0, fs), path' ∈ root0, path ∈ root0, parent(path') ∈ root0 ⊦ fid ∈ fs ↔ fid ∈ ((root0.fids \ sdt.fids) \ (path' ⊑ path ⊃ dnode(md0, ∅) -- restn(# path', path);dnode(md0, ∅)).fids) ∪ sdt.fids ∨ fid ∈ ofh.fids, path' = ε, path ⊑ path'

Rule argument

Additional information

Used simplifier rules: ⊦ dnode(md, sdt).dir?; (s)