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Proof step 6 from proof fids-exists (previous step)

⊦ ∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 = fid

User applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: {fid} abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ {fid} = ∅ ++ fid; (s)% ⊦ fid ∈ ∅ ++ fid0 ↔ fid = fid0; (s) Used terms: {fid}