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Proof step 8 from proof fid-in-fs (
previous step
)
ofh[fd] = mkfh(fid, mode, m), fd ∈ ofh, ∀ fid. fid ∈ fs ↔ fid ∈ root.fids ∨ fid ∈ ofh.fids ⊦ fid ∈ root.fids, fid ∈ fs, ∃ n. n ∈ ofh ∧ ofh[n].fid = fid
User applied rule exists right
And closed the goal!
Rule argument
pos: right 3 terms: fd abort: F discard: F computed: T
Additional information
Used simplifier rules: ⊦ mkfh(fid, mode, m).fid = fid; (s) Used terms: fd