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Proof step 10 from proof dups-eqsize-ls (previous step)

alink ∈ ls, ∀ 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

User applied rule constructor cut

And obtained the following new goals

Proof step 11:
alink = mklink(ino, str), alink ∈ ls, ∀ 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

Rule argument

var: alink

Additional information

Used simplifier rules: alink freely generated by mklink;;