Metamath Proof Explorer
Description: Intersection with class abstraction. (Contributed by Peter Mazsa, 21-Jul-2021)
|
|
Ref |
Expression |
|
Hypotheses |
abeqin.1 |
⊢ 𝐴 = ( 𝐵 ∩ 𝐶 ) |
|
|
abeqin.2 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
|
Assertion |
abeqin |
⊢ 𝐴 = { 𝑥 ∈ 𝐶 ∣ 𝜑 } |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
abeqin.1 |
⊢ 𝐴 = ( 𝐵 ∩ 𝐶 ) |
2 |
|
abeqin.2 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
3 |
2
|
ineq1i |
⊢ ( 𝐵 ∩ 𝐶 ) = ( { 𝑥 ∣ 𝜑 } ∩ 𝐶 ) |
4 |
|
dfrab2 |
⊢ { 𝑥 ∈ 𝐶 ∣ 𝜑 } = ( { 𝑥 ∣ 𝜑 } ∩ 𝐶 ) |
5 |
3 1 4
|
3eqtr4i |
⊢ 𝐴 = { 𝑥 ∈ 𝐶 ∣ 𝜑 } |