Description: Alternate proof of relopabi , which is shorter but requires an extra DV condition. (Contributed by BJ, 22-Jul-2023) (Proof modification is discouraged.) (New usage is discouraged.)
| Ref | Expression | ||
|---|---|---|---|
| Hypothesis | relopabiv.1 | ⊢ 𝐴 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } | |
| Assertion | relopabiv | ⊢ Rel 𝐴 |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | relopabiv.1 | ⊢ 𝐴 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } | |
| 2 | vex | ⊢ 𝑥 ∈ V | |
| 3 | vex | ⊢ 𝑦 ∈ V | |
| 4 | 2 3 | pm3.2i | ⊢ ( 𝑥 ∈ V ∧ 𝑦 ∈ V ) |
| 5 | 4 | a1i | ⊢ ( 𝜑 → ( 𝑥 ∈ V ∧ 𝑦 ∈ V ) ) |
| 6 | 5 | ssopab2i | ⊢ { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 } ⊆ { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ V ∧ 𝑦 ∈ V ) } |
| 7 | df-xp | ⊢ ( V × V ) = { ⟨ 𝑥 , 𝑦 ⟩ ∣ ( 𝑥 ∈ V ∧ 𝑦 ∈ V ) } | |
| 8 | 6 1 7 | 3sstr4i | ⊢ 𝐴 ⊆ ( V × V ) |
| 9 | df-rel | ⊢ ( Rel 𝐴 ↔ 𝐴 ⊆ ( V × V ) ) | |
| 10 | 8 9 | mpbir | ⊢ Rel 𝐴 |