Description: Two ways to express equality with an ordered pair. (Contributed by NM, 25-Feb-2014)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | eqop2.1 | ⊢ 𝐵 ∈ V | |
| eqop2.2 | ⊢ 𝐶 ∈ V | ||
| Assertion | eqop2 | ⊢ ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ ( 𝐴 ∈ ( V × V ) ∧ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eqop2.1 | ⊢ 𝐵 ∈ V | |
| 2 | eqop2.2 | ⊢ 𝐶 ∈ V | |
| 3 | 1 2 | opelvv | ⊢ 〈 𝐵 , 𝐶 〉 ∈ ( V × V ) |
| 4 | eleq1 | ⊢ ( 𝐴 = 〈 𝐵 , 𝐶 〉 → ( 𝐴 ∈ ( V × V ) ↔ 〈 𝐵 , 𝐶 〉 ∈ ( V × V ) ) ) | |
| 5 | 3 4 | mpbiri | ⊢ ( 𝐴 = 〈 𝐵 , 𝐶 〉 → 𝐴 ∈ ( V × V ) ) |
| 6 | eqop | ⊢ ( 𝐴 ∈ ( V × V ) → ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) ) | |
| 7 | 5 6 | biadanii | ⊢ ( 𝐴 = 〈 𝐵 , 𝐶 〉 ↔ ( 𝐴 ∈ ( V × V ) ∧ ( ( 1st ‘ 𝐴 ) = 𝐵 ∧ ( 2nd ‘ 𝐴 ) = 𝐶 ) ) ) |