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Proof step 4 from proof count-entries (previous step)

∀ str. ¬ str ∈ sdt, ∀ entries0, sdt. # entries0 < # ∅ → (∀ str. str ∈ entries0 ↔ str ∈ sdt) → # entries0 = # sdt ⊦ sdt = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ sdt1 = sdt2 ↔ (∀ str. (str ∈ sdt1 ↔ str ∈ sdt2) ∧ sdt1[str] = sdt2[str])
Proof step 6:
∀ str. ¬ str ∈ sdt, ∀ entries0, sdt. # entries0 < # ∅ → (∀ str. str ∈ entries0 ↔ str ∈ sdt) → # entries0 = # sdt ⊦ ∀ str. (str ∈ sdt ↔ str ∈ ∅) ∧ sdt[str] = (∅)[str]

Rule argument

spec: store, inst: directory-tree-inst name: Extension seq: ⊦ sdt1 = sdt2 ↔ (∀ str. (str ∈ sdt1 ↔ str ∈ sdt2) ∧ sdt1[str] = sdt2[str]) subst: [sdt2, sdt1] → [∅, sdt] flags: <all paths>

Additional information

Used terms: ∅, sdt