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Proof step 9 from proof fids-exists (previous step)

∀ 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

User applied rule cut formula

And obtained the following new goals

Proof step 10:
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
Proof step 12:
∀ 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) ⊦ sdt = ∅, ∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt.fids

Rule argument

fma: sdt = ∅

Additional information

Used formulas: sdt = ∅