Metamath Proof Explorer


Theorem alexeqg

Description: Two ways to express substitution of A for x in ph . This is the analogue for classes of sbalex . (Contributed by NM, 2-Mar-1995) (Revised by BJ, 27-Apr-2019)

Ref Expression
Assertion alexeqg ( 𝐴𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )

Proof

Step Hyp Ref Expression
1 eqeq2 ( 𝑦 = 𝐴 → ( 𝑥 = 𝑦𝑥 = 𝐴 ) )
2 1 anbi1d ( 𝑦 = 𝐴 → ( ( 𝑥 = 𝑦𝜑 ) ↔ ( 𝑥 = 𝐴𝜑 ) ) )
3 2 exbidv ( 𝑦 = 𝐴 → ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )
4 1 imbi1d ( 𝑦 = 𝐴 → ( ( 𝑥 = 𝑦𝜑 ) ↔ ( 𝑥 = 𝐴𝜑 ) ) )
5 4 albidv ( 𝑦 = 𝐴 → ( ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )
6 sbalex ( ∃ 𝑥 ( 𝑥 = 𝑦𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝑦𝜑 ) )
7 3 5 6 vtoclbg ( 𝐴𝑉 → ( ∃ 𝑥 ( 𝑥 = 𝐴𝜑 ) ↔ ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )
8 7 bicomd ( 𝐴𝑉 → ( ∀ 𝑥 ( 𝑥 = 𝐴𝜑 ) ↔ ∃ 𝑥 ( 𝑥 = 𝐴𝜑 ) ) )