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Proof step 12 from proof dups-eqsize-entries (previous step)

∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ ks0, ino, entries. # ks0 < # ks → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ # ks = 0, # ks = # entries, ks = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0)
Proof step 14:
∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∀ ks0, ino, entries. # ks0 < # ks → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ # ks = # entries, # ks = 0, ∀ key. key ∈ ks ↔ key ∈ ∅

Rule argument

spec: set-basic, inst: node-index name: Extension seq: ⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0) subst: [ks0, ks] → [∅, ks] flags: <all paths>

Additional information

Used terms: ∅, ks