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Proof step 5 from proof fid-in-fs (previous step)

fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids, ofh[fd] = mkfh(fid, mode, m), fd ∈ ofh, ∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids ⊦ fid ∈ fs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information