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Proof step 19 from proof count-entries (previous step)

str0 ∈ entries0[str, ino] ↔ str0 ∈ sdt0[str, tree], ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 = # sdt0, str0 ∈ entries0 ↔ str0 ∈ sdt0, str ∈ sdt0, str ∈ entries0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ str ∈ entries[str0, ino] ↔ str = str0 ∨ str ∈ entries
Proof step 21:
str0 = str ∨ str0 ∈ entries0 ↔ str0 ∈ sdt0[str, tree], ∀ entries1, sdt. # entries1 < # entries0[str, ino] → (∀ str. str ∈ entries1 ↔ str ∈ sdt) → # entries1 = # sdt ⊦ # entries0 = # sdt0, str0 ∈ entries0 ↔ str0 ∈ sdt0, str ∈ entries0, str ∈ sdt0

Rule argument

spec: store, inst: dir-entries name: In-insert seq: ⊦ str ∈ entries[str0, ino] ↔ str = str0 ∨ str ∈ entries subst: [ino, str0, entries, str] → [ino, str, entries0, str0] flags: <all paths>

Additional information

Used terms: ino, str, entries0, str0