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 ( 𝐴𝑉 → ( ( ∀ 𝑥 ( 𝜑𝑥 = 𝐴 ) ∧ ∀ 𝑥 ( 𝜑𝑥 = 𝐵 ) ) → 𝐴 = 𝐵 ) )

Proof

Step Hyp Ref Expression
1 bibi1 ( ( 𝜑𝑥 = 𝐴 ) → ( ( 𝜑𝑥 = 𝐵 ) ↔ ( 𝑥 = 𝐴𝑥 = 𝐵 ) ) )
2 1 biimpa ( ( ( 𝜑𝑥 = 𝐴 ) ∧ ( 𝜑𝑥 = 𝐵 ) ) → ( 𝑥 = 𝐴𝑥 = 𝐵 ) )
3 2 biimpd ( ( ( 𝜑𝑥 = 𝐴 ) ∧ ( 𝜑𝑥 = 𝐵 ) ) → ( 𝑥 = 𝐴𝑥 = 𝐵 ) )
4 3 alanimi ( ( ∀ 𝑥 ( 𝜑𝑥 = 𝐴 ) ∧ ∀ 𝑥 ( 𝜑𝑥 = 𝐵 ) ) → ∀ 𝑥 ( 𝑥 = 𝐴𝑥 = 𝐵 ) )
5 sbceqal ( 𝐴𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴𝑥 = 𝐵 ) → 𝐴 = 𝐵 ) )
6 4 5 syl5 ( 𝐴𝑉 → ( ( ∀ 𝑥 ( 𝜑𝑥 = 𝐴 ) ∧ ∀ 𝑥 ( 𝜑𝑥 = 𝐵 ) ) → 𝐴 = 𝐵 ) )