Proof of Theorem rabeqsnd
Step | Hyp | Ref
| Expression |
1 | | rabeqsnd.3 |
. . . . . 6
⊢ (((𝜑 ∧ 𝑥 ∈ 𝐴) ∧ 𝜓) → 𝑥 = 𝐵) |
2 | 1 | expl 458 |
. . . . 5
⊢ (𝜑 → ((𝑥 ∈ 𝐴 ∧ 𝜓) → 𝑥 = 𝐵)) |
3 | 2 | alrimiv 1930 |
. . . 4
⊢ (𝜑 → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → 𝑥 = 𝐵)) |
4 | | rabeqsnd.1 |
. . . . . . . 8
⊢ (𝜑 → 𝐵 ∈ 𝐴) |
5 | | rabeqsnd.2 |
. . . . . . . 8
⊢ (𝜑 → 𝜒) |
6 | 4, 5 | jca 512 |
. . . . . . 7
⊢ (𝜑 → (𝐵 ∈ 𝐴 ∧ 𝜒)) |
7 | 6 | a1d 25 |
. . . . . 6
⊢ (𝜑 → (𝑥 = 𝐵 → (𝐵 ∈ 𝐴 ∧ 𝜒))) |
8 | 7 | alrimiv 1930 |
. . . . 5
⊢ (𝜑 → ∀𝑥(𝑥 = 𝐵 → (𝐵 ∈ 𝐴 ∧ 𝜒))) |
9 | | eleq1 2826 |
. . . . . . . 8
⊢ (𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ↔ 𝐵 ∈ 𝐴)) |
10 | | rabeqsnd.0 |
. . . . . . . 8
⊢ (𝑥 = 𝐵 → (𝜓 ↔ 𝜒)) |
11 | 9, 10 | anbi12d 631 |
. . . . . . 7
⊢ (𝑥 = 𝐵 → ((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ (𝐵 ∈ 𝐴 ∧ 𝜒))) |
12 | 11 | pm5.74i 270 |
. . . . . 6
⊢ ((𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜓)) ↔ (𝑥 = 𝐵 → (𝐵 ∈ 𝐴 ∧ 𝜒))) |
13 | 12 | albii 1822 |
. . . . 5
⊢
(∀𝑥(𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜓)) ↔ ∀𝑥(𝑥 = 𝐵 → (𝐵 ∈ 𝐴 ∧ 𝜒))) |
14 | 8, 13 | sylibr 233 |
. . . 4
⊢ (𝜑 → ∀𝑥(𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜓))) |
15 | 3, 14 | jca 512 |
. . 3
⊢ (𝜑 → (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → 𝑥 = 𝐵) ∧ ∀𝑥(𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜓)))) |
16 | | albiim 1892 |
. . 3
⊢
(∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ 𝑥 = 𝐵) ↔ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) → 𝑥 = 𝐵) ∧ ∀𝑥(𝑥 = 𝐵 → (𝑥 ∈ 𝐴 ∧ 𝜓)))) |
17 | 15, 16 | sylibr 233 |
. 2
⊢ (𝜑 → ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ 𝑥 = 𝐵)) |
18 | | rabeqsn 4602 |
. 2
⊢ ({𝑥 ∈ 𝐴 ∣ 𝜓} = {𝐵} ↔ ∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜓) ↔ 𝑥 = 𝐵)) |
19 | 17, 18 | sylibr 233 |
1
⊢ (𝜑 → {𝑥 ∈ 𝐴 ∣ 𝜓} = {𝐵}) |