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Proof step 36 from proof rename-fids (previous step)

root0[path'] = fnode(fid), fnode(fid1) = root0[path], (root0 -- path)[parent(path')].dir?, root0[parent(path')].dir?, fids-cons(root0, fs, ofh), py(user, parent(path), root0, fs), py(user, parent(path'), root0, fs), fid0 ∈ root0.fids, path' ∈ root0, path ∈ root0, parent(path') ∈ root0, fid0 ∈ fs, path' ⊑ path ⊦ fid = fid1, fid0 = fid, fid0 = fid1, path' = ε, fid0 ∈ ofh.fids, fid0 ∈ (root0.fids \ {fid1} ) \ (path' ⊑ path ⊃ fnode(fid) -- restn(# path', path);fnode(fid)).fids, fid ∈ ofh.fids, fid ∈ (root0.fids \ {fid1} ) \ (fnode(fid) -- restn(# path', path)).fids, path ⊑ path'

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
root0[path'] = fnode(fid), fnode(fid1) = root0[path], root0[parent(path')].dir?, (root0 -- path)[parent(path')].dir?, fids-cons(root0, fs, ofh), py(user, parent(path'), root0, fs), py(user, parent(path), root0, fs), fid0 ∈ fs, parent(path') ∈ root0, path ∈ root0, path' ∈ root0, fid0 ∈ root0.fids, path' ⊑ path ⊦ fid = fid1, fid0 = fid1, fid0 = fid, path' = ε, path' = path, fid0 ∈ ofh.fids, fid ∈ (root0.fids \ {fid1} ) \ (fnode(fid) -- restn(# path', path)).fids, fid ∈ ofh.fids, fid0 ∈ (root0.fids \ {fid1} ) \ (fnode(fid) -- restn(# path', path)).fids

Rule argument

Additional information

Used simplifier rules: ⊦ p ⊑ p0 → (p0 ⊑ p ↔ p = p0); (s)