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Proof step 15 from proof inv-unlink (previous step)

root0[path] = fnode(fid), root0.dir?, fids-cons(root0, fs, ofh), py(user, parent(path), root0, fs), path ∈ root0 ⊦ path = ε, fids-cons(root0 -- path, fs -- fid, ofh), fid ∈ ofh.fids, fid ∈ root0.fids \ {fid}

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
⊦ 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 17:
fid0 ∈ fs ↔ fid0 ∈ root0.fids ∨ fid0 ∈ ofh.fids, root0[path] = fnode(fid), root0.dir?, fids-cons(root0, fs, ofh), py(user, parent(path), root0, fs), path ∈ root0 ⊦ fid0 ≠ fid ∧ fid0 ∈ fs ↔ fid0 ∈ root0.fids \ {fid} ∨ fid0 ∈ ofh.fids, path = ε, fid ∈ ofh.fids, fid ∈ root0.fids \ {fid}

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)));% ⊦ fnode(fid).fids = {fid} ; (s)%p ≠ ε ⊦ p ∈ tree → (tree -- p).fids = tree.fids \ tree[p].fids; (s)% ⊦ fid ∈ fs -- fid0 ↔ fid ≠ fid0 ∧ fid ∈ fs; (s)