Step | Hyp | Ref
| Expression |
1 | | disj1 4385 |
. 2
⊢ ((𝐴 ∩ {𝐵}) = ∅ ↔ ∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ {𝐵})) |
2 | | con2b 360 |
. . . 4
⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ {𝐵}) ↔ (𝑥 ∈ {𝐵} → ¬ 𝑥 ∈ 𝐴)) |
3 | | velsn 4578 |
. . . . 5
⊢ (𝑥 ∈ {𝐵} ↔ 𝑥 = 𝐵) |
4 | 3 | imbi1i 350 |
. . . 4
⊢ ((𝑥 ∈ {𝐵} → ¬ 𝑥 ∈ 𝐴) ↔ (𝑥 = 𝐵 → ¬ 𝑥 ∈ 𝐴)) |
5 | | imnan 400 |
. . . 4
⊢ ((𝑥 = 𝐵 → ¬ 𝑥 ∈ 𝐴) ↔ ¬ (𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴)) |
6 | 2, 4, 5 | 3bitri 297 |
. . 3
⊢ ((𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ {𝐵}) ↔ ¬ (𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴)) |
7 | 6 | albii 1822 |
. 2
⊢
(∀𝑥(𝑥 ∈ 𝐴 → ¬ 𝑥 ∈ {𝐵}) ↔ ∀𝑥 ¬ (𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴)) |
8 | | alnex 1784 |
. . 3
⊢
(∀𝑥 ¬
(𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴) ↔ ¬ ∃𝑥(𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴)) |
9 | | dfclel 2818 |
. . 3
⊢ (𝐵 ∈ 𝐴 ↔ ∃𝑥(𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴)) |
10 | 8, 9 | xchbinxr 335 |
. 2
⊢
(∀𝑥 ¬
(𝑥 = 𝐵 ∧ 𝑥 ∈ 𝐴) ↔ ¬ 𝐵 ∈ 𝐴) |
11 | 1, 7, 10 | 3bitri 297 |
1
⊢ ((𝐴 ∩ {𝐵}) = ∅ ↔ ¬ 𝐵 ∈ 𝐴) |