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Proof step 9 from proof dups-eqsize-entries (previous step)

∀ key. key.dentry? ∧ key.ino = ino → ¬ key.name ∈ entries, ∀ ks0, ino, entries. # ks0 < # ∅ → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ ∀ str. (str ∈ entries ↔ str ∈ ∅) ∧ (str ∈ entries → entries[str] = (∅)[str])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
str ∈ entries, ∀ key. key.dentry? ∧ key.ino = ino → ¬ key.name ∈ entries, ∀ ks0, ino, entries. # ks0 < # ∅ → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ str ∈ ∅; (s)