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Proof step 14 from proof dups-eqsize-ls (previous step)

∀ key. key ∈ ks → key.dentry?, ∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls), ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ks = 0, # ks = # ls, ks = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ ks = ks0 ↔ (∀ key. key ∈ ks ↔ key ∈ ks0)
Proof step 16:
∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls), ∀ key. key ∈ ks → key.dentry?, ∀ ls0, ks. # ls0 < # ls → (∀ key. key ∈ ks → key.dentry?) ∧ (∀ key. key.dentry? → (key ∈ ks ↔ mklink(key.ino, key.name) ∈ ls0)) → # ks = # ls0 ⊦ # ks = # ls, # 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