Step |
Hyp |
Ref |
Expression |
1 |
|
eqreu.1 |
⊢ ( 𝑥 = 𝐵 → ( 𝜑 ↔ 𝜓 ) ) |
2 |
|
ralbiim |
⊢ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝑥 = 𝐵 ) ↔ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ ∀ 𝑥 ∈ 𝐴 ( 𝑥 = 𝐵 → 𝜑 ) ) ) |
3 |
1
|
ceqsralv |
⊢ ( 𝐵 ∈ 𝐴 → ( ∀ 𝑥 ∈ 𝐴 ( 𝑥 = 𝐵 → 𝜑 ) ↔ 𝜓 ) ) |
4 |
3
|
anbi2d |
⊢ ( 𝐵 ∈ 𝐴 → ( ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ ∀ 𝑥 ∈ 𝐴 ( 𝑥 = 𝐵 → 𝜑 ) ) ↔ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ 𝜓 ) ) ) |
5 |
2 4
|
bitrid |
⊢ ( 𝐵 ∈ 𝐴 → ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝑥 = 𝐵 ) ↔ ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ 𝜓 ) ) ) |
6 |
|
reu6i |
⊢ ( ( 𝐵 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝑥 = 𝐵 ) ) → ∃! 𝑥 ∈ 𝐴 𝜑 ) |
7 |
6
|
ex |
⊢ ( 𝐵 ∈ 𝐴 → ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 ↔ 𝑥 = 𝐵 ) → ∃! 𝑥 ∈ 𝐴 𝜑 ) ) |
8 |
5 7
|
sylbird |
⊢ ( 𝐵 ∈ 𝐴 → ( ( ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ 𝜓 ) → ∃! 𝑥 ∈ 𝐴 𝜑 ) ) |
9 |
8
|
3impib |
⊢ ( ( 𝐵 ∈ 𝐴 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ∧ 𝜓 ) → ∃! 𝑥 ∈ 𝐴 𝜑 ) |
10 |
9
|
3com23 |
⊢ ( ( 𝐵 ∈ 𝐴 ∧ 𝜓 ∧ ∀ 𝑥 ∈ 𝐴 ( 𝜑 → 𝑥 = 𝐵 ) ) → ∃! 𝑥 ∈ 𝐴 𝜑 ) |