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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ sdt = ∅ ↔ ¬(∃ str, tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 14:
∀ 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) ⊦ ∀ str, tree, sdt0. ¬(sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0), ∃ fids. ∀ fid0. fid0 ∈ fids ↔ fid0 ∈ sdt.fids

Rule argument

spec: store, inst: directory-tree-inst name: exnotempty seq: ⊦ sdt = ∅ ↔ ¬(∃ str, tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0) subst: [sdt] → [sdt] flags: <all paths>

Additional information

Used terms: sdt