Step | Hyp | Ref
| Expression |
1 | | elxp 5611 |
. 2
⊢ (𝐴 ∈ (𝐵 × 𝐶) ↔ ∃𝑏∃𝑐(𝐴 = 〈𝑏, 𝑐〉 ∧ (𝑏 ∈ 𝐵 ∧ 𝑐 ∈ 𝐶))) |
2 | | vex 3434 |
. . . . . . 7
⊢ 𝑏 ∈ V |
3 | | vex 3434 |
. . . . . . 7
⊢ 𝑐 ∈ V |
4 | 2, 3 | op1std 7827 |
. . . . . 6
⊢ (𝐴 = 〈𝑏, 𝑐〉 → (1st ‘𝐴) = 𝑏) |
5 | 4 | eleq1d 2824 |
. . . . 5
⊢ (𝐴 = 〈𝑏, 𝑐〉 → ((1st ‘𝐴) ∈ 𝐵 ↔ 𝑏 ∈ 𝐵)) |
6 | 5 | biimpar 477 |
. . . 4
⊢ ((𝐴 = 〈𝑏, 𝑐〉 ∧ 𝑏 ∈ 𝐵) → (1st ‘𝐴) ∈ 𝐵) |
7 | 6 | adantrr 713 |
. . 3
⊢ ((𝐴 = 〈𝑏, 𝑐〉 ∧ (𝑏 ∈ 𝐵 ∧ 𝑐 ∈ 𝐶)) → (1st ‘𝐴) ∈ 𝐵) |
8 | 7 | exlimivv 1938 |
. 2
⊢
(∃𝑏∃𝑐(𝐴 = 〈𝑏, 𝑐〉 ∧ (𝑏 ∈ 𝐵 ∧ 𝑐 ∈ 𝐶)) → (1st ‘𝐴) ∈ 𝐵) |
9 | 1, 8 | sylbi 216 |
1
⊢ (𝐴 ∈ (𝐵 × 𝐶) → (1st ‘𝐴) ∈ 𝐵) |