Metamath Proof Explorer
Description: Equality of restricted class abstractions. Deduction form of rabeq .
(Contributed by Glauco Siliprandi, 5-Apr-2020)
|
|
Ref |
Expression |
|
Hypothesis |
rabeqdv.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
Assertion |
rabeqdv |
⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
rabeqdv.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
rabeq |
⊢ ( 𝐴 = 𝐵 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |
3 |
1 2
|
syl |
⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |