Metamath Proof Explorer


Theorem r19.12sn

Description: Special case of r19.12 where its converse holds. (Contributed by NM, 19-May-2008) (Revised by Mario Carneiro, 23-Apr-2015) (Revised by BJ, 18-Mar-2020)

Ref Expression
Assertion r19.12sn ( 𝐴𝑉 → ( ∃ 𝑥 ∈ { 𝐴 } ∀ 𝑦𝐵 𝜑 ↔ ∀ 𝑦𝐵𝑥 ∈ { 𝐴 } 𝜑 ) )

Proof

Step Hyp Ref Expression
1 sbcralg ( 𝐴𝑉 → ( [ 𝐴 / 𝑥 ]𝑦𝐵 𝜑 ↔ ∀ 𝑦𝐵 [ 𝐴 / 𝑥 ] 𝜑 ) )
2 rexsns ( ∃ 𝑥 ∈ { 𝐴 } ∀ 𝑦𝐵 𝜑[ 𝐴 / 𝑥 ]𝑦𝐵 𝜑 )
3 rexsns ( ∃ 𝑥 ∈ { 𝐴 } 𝜑[ 𝐴 / 𝑥 ] 𝜑 )
4 3 ralbii ( ∀ 𝑦𝐵𝑥 ∈ { 𝐴 } 𝜑 ↔ ∀ 𝑦𝐵 [ 𝐴 / 𝑥 ] 𝜑 )
5 1 2 4 3bitr4g ( 𝐴𝑉 → ( ∃ 𝑥 ∈ { 𝐴 } ∀ 𝑦𝐵 𝜑 ↔ ∀ 𝑦𝐵𝑥 ∈ { 𝐴 } 𝜑 ) )