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Proof step 18 from proof rename-fids (previous step)

fid0 ∈ fs ↔ fid0 ∈ root0.fids ∨ fid0 ∈ ofh.fids, root0[path'] = fnode(fid), root0[parent(path')].dir?, fids-cons(root0, fs, ofh), py(user, parent(path'), root0, fs), py(user, parent(path), root0, fs), parent(path') ∈ root0, path ∈ root0, path' ∈ root0 ⊦ fid0 ≠ fid ∧ fid0 ∈ fs ↔ fid0 ∈ (root0 -- path)[path', root0[path]].fids ∨ fid0 ∈ ofh.fids, path' = ε, root0[path].dir?, fid ∈ (root0 -- path)[path', root0[path]].fids, fid ∈ ofh.fids, path ⊑ path', path' ≠ ε ∧ path' ∈ root0 -- path ∧ (root0 -- path)[parent(path')].dir?

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
fid0 ∈ fs ↔ fid0 ∈ root0.fids ∨ fid0 ∈ ofh.fids, root0[path'] = fnode(fid), root0[parent(path')].dir?, fids-cons(root0, fs, ofh), py(user, parent(path), root0, fs), py(user, parent(path'), root0, fs), path' ∈ root0, path ∈ root0, parent(path') ∈ root0 ⊦ fid0 ≠ fid ∧ fid0 ∈ fs ↔ fid0 ∈ (root0 -- path)[path', root0[path]].fids ∨ fid0 ∈ ofh.fids, path' = ε, root0[path].dir?, (root0 -- path)[parent(path')].dir?, fid ∈ ofh.fids, fid ∈ (root0 -- path)[path', root0[path]].fids, path ⊑ path'

Rule argument

Additional information

Used simplifier rules: p ≠ ε ⊦ p ∈ tree → (q ∈ tree -- p ↔ ¬ p ⊑ q ∧ q ∈ tree); (s)% ⊦ p ≠ ε → (parent(p) = ε ↔ p = p.head '); (s)% ⊦ ¬ tree.dir? → (p ∈ tree ↔ p = ε); (s)% ⊦ ε ∈ tree; (s)% ⊦ tree[ε] = tree; (s)