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Description: A way to define an ordered-pair class abstraction without using existential quantifiers. (Contributed by NM, 18-Aug-2006) (Revised by Mario Carneiro, 31-Aug-2015)
| Ref | Expression | ||
|---|---|---|---|
| Assertion | dfopab2 | ⊢ { 〈 𝑥 , 𝑦 〉 ∣ 𝜑 } = { 𝑧 ∈ ( V × V ) ∣ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 } |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | nfsbc1v | ⊢ Ⅎ 𝑥 [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 | |
| 2 | 1 | 19.41 | ⊢ ( ∃ 𝑥 ( ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ↔ ( ∃ 𝑥 ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 3 | sbcopeq1a | ⊢ ( 𝑧 = 〈 𝑥 , 𝑦 〉 → ( [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ↔ 𝜑 ) ) | |
| 4 | 3 | pm5.32i | ⊢ ( ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ↔ ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) ) |
| 5 | 4 | exbii | ⊢ ( ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ↔ ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) ) |
| 6 | nfcv | ⊢ Ⅎ 𝑦 ( 1st ‘ 𝑧 ) | |
| 7 | nfsbc1v | ⊢ Ⅎ 𝑦 [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 | |
| 8 | 6 7 | nfsbcw | ⊢ Ⅎ 𝑦 [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 |
| 9 | 8 | 19.41 | ⊢ ( ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ↔ ( ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 10 | 5 9 | bitr3i | ⊢ ( ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) ↔ ( ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 11 | 10 | exbii | ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) ↔ ∃ 𝑥 ( ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 12 | elvv | ⊢ ( 𝑧 ∈ ( V × V ) ↔ ∃ 𝑥 ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ) | |
| 13 | 12 | anbi1i | ⊢ ( ( 𝑧 ∈ ( V × V ) ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ↔ ( ∃ 𝑥 ∃ 𝑦 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 14 | 2 11 13 | 3bitr4i | ⊢ ( ∃ 𝑥 ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) ↔ ( 𝑧 ∈ ( V × V ) ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) ) |
| 15 | 14 | abbii | ⊢ { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) } = { 𝑧 ∣ ( 𝑧 ∈ ( V × V ) ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) } |
| 16 | df-opab | ⊢ { 〈 𝑥 , 𝑦 〉 ∣ 𝜑 } = { 𝑧 ∣ ∃ 𝑥 ∃ 𝑦 ( 𝑧 = 〈 𝑥 , 𝑦 〉 ∧ 𝜑 ) } | |
| 17 | df-rab | ⊢ { 𝑧 ∈ ( V × V ) ∣ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 } = { 𝑧 ∣ ( 𝑧 ∈ ( V × V ) ∧ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 ) } | |
| 18 | 15 16 17 | 3eqtr4i | ⊢ { 〈 𝑥 , 𝑦 〉 ∣ 𝜑 } = { 𝑧 ∈ ( V × V ) ∣ [ ( 1st ‘ 𝑧 ) / 𝑥 ] [ ( 2nd ‘ 𝑧 ) / 𝑦 ] 𝜑 } |