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Proof step 11 from proof fids-exists (
previous step
)
∀ sdt0. # sdt0 < # ∅ → (∀ str. str ∈ sdt0 → (∃ fids. ∀ fid. fid ∈ fids ↔ fid ∈ sdt0[str].fids)) → (∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt0.fids) ⊦ ∃ fids. ∀ fid. ¬ fid ∈ fids
User applied rule exists right
And closed the goal!
Rule argument
pos: right 1 terms: ∅ abort: F discard: F computed: F
Additional information
Used simplifier rules: ⊦ ¬ fid ∈ ∅; (s) Used terms: ∅