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Proof step 15 from proof dups-eqsize-entries (previous step)

key ∈ ks, ∀ 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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ key ∈ ks ↔ (∃ ks0. ks = ks0 ++ key ∧ ¬ key ∈ ks0)
Proof step 17:
∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∃ ks0. ks = ks0 ++ key ∧ ¬ key ∈ ks0, ∀ 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

Rule argument

spec: set-basic, inst: node-index name: indef seq: ⊦ key ∈ ks ↔ (∃ ks0. ks = ks0 ++ key ∧ ¬ key ∈ ks0) subst: [ks, key] → [ks, key] flags: <all paths>

Additional information

Used terms: ks, key