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Proof step 20 from proof ofh-fids-new (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: fd

Additional information

Used simplifier rules: ⊦ fd ∈ ofh[fd, fh]; (s)% ⊦ ofh[fd, fh][fd] = fh; (s)% ⊦ mkfh(fid, mode, m).fid = fid; (s) Used terms: fd