Description: Two ways to express equality with an ordered pair. (Contributed by NM, 3-Sep-2007) (Proof shortened by Mario Carneiro, 26-Apr-2015)
Ref | Expression | ||
---|---|---|---|
Assertion | eqop | ⊢ ( 𝐴 ∈ ( 𝑉 × 𝑊 ) → ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 1st2nd2 | ⊢ ( 𝐴 ∈ ( 𝑉 × 𝑊 ) → 𝐴 = 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 ) | |
2 | 1 | eqeq1d | ⊢ ( 𝐴 ∈ ( 𝑉 × 𝑊 ) → ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 = 〈 𝐵 , 𝐶 〉 ) ) |
3 | fvex | ⊢ ( 1st ‘ 𝐴 ) ∈ V | |
4 | fvex | ⊢ ( 2nd ‘ 𝐴 ) ∈ V | |
5 | 3 4 | opth | ⊢ ( 〈 ( 1st ‘ 𝐴 ) , ( 2nd ‘ 𝐴 ) 〉 = 〈 𝐵 , 𝐶 〉 ↔ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) |
6 | 2 5 | bitrdi | ⊢ ( 𝐴 ∈ ( 𝑉 × 𝑊 ) → ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) ) |