Metamath Proof Explorer
Description: Deduction from a wff to a class abstraction. (Contributed by NM, 9-Jul-1994) (Proof shortened by Wolf Lammen, 16-Nov-2019)
|
|
Ref |
Expression |
|
Hypothesis |
abbi1dv.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝑥 ∈ 𝐴 ) ) |
|
Assertion |
abbi1dv |
⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = 𝐴 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
abbi1dv.1 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝑥 ∈ 𝐴 ) ) |
2 |
1
|
bicomd |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝜓 ) ) |
3 |
2
|
abbi2dv |
⊢ ( 𝜑 → 𝐴 = { 𝑥 ∣ 𝜓 } ) |
4 |
3
|
eqcomd |
⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = 𝐴 ) |