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Proof step 6 from proof dups-eqsize-ls (previous step)

ks = ∅, ∀ 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 = # ls

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ 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 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ ls ≠ ∅ → # ls ≠ 0; (forward) from specification set-basic[alink-set]%size-zero: ⊦ # ls ≠ 0 → ls ≠ ∅; (forward) from specification set-basic[alink-set]% ⊦ ¬ key ∈ ∅; (s)% ⊦ # ∅ = 0; (s)