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Proof step 5 from proof dups-eqsize-entries (previous step)

∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∀ ks0, ino, entries. # ks0 < # ks → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ # ks = # entries

User applied rule cut formula

And obtained the following new goals

Proof step 6:
ks = ∅, ∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ ks0, ino, entries. # ks0 < # ks → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ # ks = # entries
Proof step 11:
∀ key. key ∈ ks → key.dentry? ∧ key.ino = ino, ∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks ↔ key.name ∈ entries), ∀ ks0, ino, entries. # ks0 < # ks → (∀ key. key ∈ ks0 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks0 ↔ key.name ∈ entries)) → # ks0 = # entries ⊦ # ks = # entries, ks = ∅

Rule argument

fma: ks = ∅

Additional information

Used formulas: ks = ∅