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Proof step 14 from proof ofh-fids-new (previous step)

fid = fh.fid ∨ (∃ n. n ∈ ofh ∧ ofh[n].fid = fid) ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, fh] ∧ ofh[fd, fh][n].fid = fid

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information