Metamath Proof Explorer


Theorem iinxsng

Description: A singleton index picks out an instance of an indexed intersection's argument. (Contributed by NM, 15-Jan-2012) (Proof shortened by Mario Carneiro, 17-Nov-2016)

Ref Expression
Hypothesis iinxsng.1 ( 𝑥 = 𝐴𝐵 = 𝐶 )
Assertion iinxsng ( 𝐴𝑉 𝑥 ∈ { 𝐴 } 𝐵 = 𝐶 )

Proof

Step Hyp Ref Expression
1 iinxsng.1 ( 𝑥 = 𝐴𝐵 = 𝐶 )
2 df-iin 𝑥 ∈ { 𝐴 } 𝐵 = { 𝑦 ∣ ∀ 𝑥 ∈ { 𝐴 } 𝑦𝐵 }
3 1 eleq2d ( 𝑥 = 𝐴 → ( 𝑦𝐵𝑦𝐶 ) )
4 3 ralsng ( 𝐴𝑉 → ( ∀ 𝑥 ∈ { 𝐴 } 𝑦𝐵𝑦𝐶 ) )
5 4 abbi1dv ( 𝐴𝑉 → { 𝑦 ∣ ∀ 𝑥 ∈ { 𝐴 } 𝑦𝐵 } = 𝐶 )
6 2 5 eqtrid ( 𝐴𝑉 𝑥 ∈ { 𝐴 } 𝐵 = 𝐶 )