Metamath Proof Explorer

Theorem sbeqalb

Description: Theorem *14.121 in WhiteheadRussell p. 185. (Contributed by Andrew Salmon, 28-Jun-2011) (Proof shortened by Wolf Lammen, 9-May-2013)

Ref Expression
Assertion sbeqalb ⊢ A ∈ V → ∀ x φ ↔ x = A ∧ ∀ x φ ↔ x = B → A = B

Proof

Step Hyp Ref Expression
1 bibi1 ⊢ φ ↔ x = A → φ ↔ x = B ↔ x = A ↔ x = B
2 1 biimpa ⊢ φ ↔ x = A ∧ φ ↔ x = B → x = A ↔ x = B
3 2 biimpd ⊢ φ ↔ x = A ∧ φ ↔ x = B → x = A → x = B
4 3 alanimi ⊢ ∀ x φ ↔ x = A ∧ ∀ x φ ↔ x = B → ∀ x x = A → x = B
5 sbceqal ⊢ A ∈ V → ∀ x x = A → x = B → A = B
6 4 5 syl5 ⊢ A ∈ V → ∀ x φ ↔ x = A ∧ ∀ x φ ↔ x = B → A = B