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Proof step 3 from proof fids-strpath (previous step)

∀ sdt0, fid. # sdt0 < # sdt → true → (fid ∈ sdt0.fids ↔ (∃ str, p. str ∈ sdt0 ∧ p ∈ sdt0[str] ∧ sdt0[str][p] = fnode(fid))) ⊦ fid ∈ sdt.fids ↔ (∃ str, p. str ∈ sdt ∧ p ∈ sdt[str] ∧ sdt[str][p] = fnode(fid))

User applied rule cut formula

And obtained the following new goals

Proof step 4:
⊦ ∅.fids = ∅
Proof step 5:
∀ sdt0, fid. # sdt0 < # sdt → true → (fid ∈ sdt0.fids ↔ (∃ str, p. str ∈ sdt0 ∧ p ∈ sdt0[str] ∧ sdt0[str][p] = fnode(fid))) ⊦ fid ∈ sdt.fids ↔ (∃ str, p. str ∈ sdt ∧ p ∈ sdt[str] ∧ sdt[str][p] = fnode(fid)), sdt = ∅

Rule argument

fma: sdt = ∅

Additional information

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