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Proof step 10 from proof fids-exists (previous step)

sdt = ∅, ∀ str. str ∈ sdt → (∃ fids. ∀ fid. fid ∈ fids ↔ fid ∈ sdt[str].fids), ∀ sdt0. # sdt0 < # sdt → (∀ str. str ∈ sdt0 → (∃ fids. ∀ fid. fid ∈ fids ↔ fid ∈ sdt0[str].fids)) → (∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt0.fids) ⊦ ∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt.fids

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ sdt0. # sdt0 < # ∅ → (∀ str. str ∈ sdt0 → (∃ fids. ∀ fid. fid ∈ fids ↔ fid ∈ sdt0[str].fids)) → (∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt0.fids) ⊦ ∃ fids. ∀ fid. ¬ fid ∈ fids

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ fid ∈ ∅; (s)% ⊦ ∅.fids = ∅; (s)% ⊦ ¬ str ∈ ∅; (s)