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Proof step 7 from proof dups-eqsize-ls (previous step)

∀ key. key.dentry? → ¬ 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 ⊦ 0 = # ls, ls = ∅

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0)
Proof step 9:
∀ key. key.dentry? → ¬ 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 ⊦ 0 = # ls, ∀ alink. alink ∈ ls ↔ alink ∈ ∅

Rule argument

spec: set-basic, inst: alink-set name: Extension seq: ⊦ ls = ls0 ↔ (∀ alink. alink ∈ ls ↔ alink ∈ ls0) subst: [ls0, ls] → [∅, ls] flags: <all paths>

Additional information

Used terms: ∅, ls