Metamath Proof Explorer


Theorem elprg

Description: A member of a pair of classes is one or the other of them, and conversely as soon as it is a set. Exercise 1 of TakeutiZaring p. 15, generalized. (Contributed by NM, 13-Sep-1995)

Ref Expression
Assertion elprg ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝐵 , 𝐶 } ↔ ( 𝐴 = 𝐵 ∨ 𝐴 = 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 eqeq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 = 𝐵 ↔ 𝐴 = 𝐵 ) )
2 eqeq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 = 𝐶 ↔ 𝐴 = 𝐶 ) )
3 1 2 orbi12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 = 𝐵 ∨ 𝑥 = 𝐶 ) ↔ ( 𝐴 = 𝐵 ∨ 𝐴 = 𝐶 ) ) )
4 dfpr2 ⊢ { 𝐵 , 𝐶 } = { 𝑥 ∣ ( 𝑥 = 𝐵 ∨ 𝑥 = 𝐶 ) }
5 3 4 elab2g ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝐵 , 𝐶 } ↔ ( 𝐴 = 𝐵 ∨ 𝐴 = 𝐶 ) ) )