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Proof step 19 from proof ofh-fids-new (previous step)

ofh[n] = mkfh(fid1, mode0, m0), fid = fid0 ⊦ fd ∈ ofh, ∃ n. n ∈ ofh[fd, mkfh(fid0, mode, m)] ∧ ofh[fd, mkfh(fid0, mode, m)][n].fid = fid

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information