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Proof step 13 from proof count-entries (previous step)

str ∈ sdt, ∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt, ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 + 1 = # sdt, str ∈ entries0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ str ∈ sdt ↔ (∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0)
Proof step 15:
∀ str0. str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt, ∃ tree, sdt0. sdt = sdt0[str, tree] ∧ ¬ str ∈ sdt0, ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 + 1 = # sdt, str ∈ entries0

Rule argument

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

Additional information

Used terms: sdt, str