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Proof step 17 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. ks = ks0 ++ key ∧ ¬ key ∈ ks0, ∀ 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 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
∀ key0. key0 = key ∨ key0 ∈ ks0 → key0.dentry? ∧ key0.ino = ino, ∀ key0. key0.dentry? ∧ key0.ino = ino → (key0 = key ∨ key0 ∈ ks0 ↔ key0.name ∈ entries), ∀ ks1, ino, entries. # ks1 < #(ks0 ++ key) → (∀ key. key ∈ ks1 → key.dentry? ∧ key.ino = ino) ∧ (∀ key. key.dentry? ∧ key.ino = ino → (key ∈ ks1 ↔ key.name ∈ entries)) → # ks1 = # entries ⊦ # ks0 + 1 = # entries, key ∈ ks0

Rule argument

Additional information

Used simplifier rules: ⊦ key ∈ ks ++ key0 ↔ key = key0 ∨ key ∈ ks; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)% ⊦ ¬ key ∈ ks → #(ks ++ key) = # ks + 1; (s)