Metamath Proof Explorer


Theorem 2nd1st

Description: Swap the members of an ordered pair. (Contributed by NM, 31-Dec-2014)

Ref Expression
Assertion 2nd1st ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → { 𝐴 } = ⟨ ( 2nd𝐴 ) , ( 1st𝐴 ) ⟩ )

Proof

Step Hyp Ref Expression
1 1st2nd2 ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ )
2 1 sneqd ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → { 𝐴 } = { ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ } )
3 2 cnveqd ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → { 𝐴 } = { ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ } )
4 3 unieqd ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → { 𝐴 } = { ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ } )
5 opswap { ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ } = ⟨ ( 2nd𝐴 ) , ( 1st𝐴 ) ⟩
6 4 5 eqtrdi ( 𝐴 ∈ ( 𝐵 × 𝐶 ) → { 𝐴 } = ⟨ ( 2nd𝐴 ) , ( 1st𝐴 ) ⟩ )