Metamath Proof Explorer


Theorem elxp6

Description: Membership in a Cartesian product. This version requires no quantifiers or dummy variables. See also elxp4 . (Contributed by NM, 9-Oct-2004)

Ref Expression
Assertion elxp6 ( 𝐴 ∈ ( 𝐵 × 𝐶 ) ↔ ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∧ ( ( 1st𝐴 ) ∈ 𝐵 ∧ ( 2nd𝐴 ) ∈ 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 elxp4 ( 𝐴 ∈ ( 𝐵 × 𝐶 ) ↔ ( 𝐴 = ⟨ dom { 𝐴 } , ran { 𝐴 } ⟩ ∧ ( dom { 𝐴 } ∈ 𝐵 ran { 𝐴 } ∈ 𝐶 ) ) )
2 1stval ( 1st𝐴 ) = dom { 𝐴 }
3 2ndval ( 2nd𝐴 ) = ran { 𝐴 }
4 2 3 opeq12i ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ = ⟨ dom { 𝐴 } , ran { 𝐴 } ⟩
5 4 eqeq2i ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ↔ 𝐴 = ⟨ dom { 𝐴 } , ran { 𝐴 } ⟩ )
6 2 eleq1i ( ( 1st𝐴 ) ∈ 𝐵 dom { 𝐴 } ∈ 𝐵 )
7 3 eleq1i ( ( 2nd𝐴 ) ∈ 𝐶 ran { 𝐴 } ∈ 𝐶 )
8 6 7 anbi12i ( ( ( 1st𝐴 ) ∈ 𝐵 ∧ ( 2nd𝐴 ) ∈ 𝐶 ) ↔ ( dom { 𝐴 } ∈ 𝐵 ran { 𝐴 } ∈ 𝐶 ) )
9 5 8 anbi12i ( ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∧ ( ( 1st𝐴 ) ∈ 𝐵 ∧ ( 2nd𝐴 ) ∈ 𝐶 ) ) ↔ ( 𝐴 = ⟨ dom { 𝐴 } , ran { 𝐴 } ⟩ ∧ ( dom { 𝐴 } ∈ 𝐵 ran { 𝐴 } ∈ 𝐶 ) ) )
10 1 9 bitr4i ( 𝐴 ∈ ( 𝐵 × 𝐶 ) ↔ ( 𝐴 = ⟨ ( 1st𝐴 ) , ( 2nd𝐴 ) ⟩ ∧ ( ( 1st𝐴 ) ∈ 𝐵 ∧ ( 2nd𝐴 ) ∈ 𝐶 ) ) )