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 = ⟨ ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) ⟩ ∧ ( ( 1st ‘ 𝐴 ) ∈ 𝐵 ∧ ( 2nd ‘ 𝐴 ) ∈ 𝐶 ) ) )

Proof

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