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Proof step 14 from proof rename-fids (previous step)

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 ⊦ path' = ε, root0[path].dir?, fids-cons((root0 -- path)[path', root0[path]], fs -- fid, ofh), fid ∈ (root0 -- path)[path', root0[path]].fids, fid ∈ ofh.fids, path ⊑ path'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ fids-cons(tree0, fs0, ofh0) → (fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)))
Proof step 16:
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?, fid ∈ ofh.fids, fid ∈ (root0 -- path)[path', root0[path]].fids, path ⊑ path'

Rule argument

Additional information

Used simplifier rules: fids-cons-reduce: ⊦ fids-cons(tree0, fs0, ofh0) → ( fids-cons(tree1, fs1, ofh1) ↔ (∀ fid. (fid ∈ fs0 ↔ fid ∈ tree0.fids ∨ fid ∈ ofh0.fids) → (fid ∈ fs1 ↔ fid ∈ tree1.fids ∨ fid ∈ ofh1.fids)));% ⊦ fid ∈ fs -- fid0 ↔ fid ≠ fid0 ∧ fid ∈ fs; (s)