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Proof step 22 from proof ofh-fids-new (previous step)

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

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: n abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ fd ∈ ofh ∧ ¬ fd0 ∈ ofh → ofh[fd0, fh][fd] = ofh[fd]; (s)% ⊦ mkfh(fid, mode, m).fid = fid; (s)% ⊦ fd ∈ ofh[fd0, fh] ↔ fd = fd0 ∨ fd ∈ ofh; (ws) Used terms: n