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Proof step 10 from proof ofh-fids-new (previous step)

fh = mkfh(fid, mode, m), n ∈ ofh[fd, fh] ⊦ ofh[fd, fh][n].fid = fid, fd ∈ ofh, ∃ n0. n0 ∈ ofh ∧ ofh[n0].fid = ofh[fd, fh][n].fid

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information