Metamath Proof Explorer
Description: Equality of restricted class abstractions. (Contributed by Jeff Madsen, 1-Dec-2009)
|
|
Ref |
Expression |
|
Hypotheses |
rabeqbidv.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
|
rabeqbidv.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
|
Assertion |
rabeqbidv |
⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
rabeqbidv.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
rabeqbidv.2 |
⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) ) |
3 |
2
|
adantr |
⊢ ( ( 𝜑 ∧ 𝑥 ∈ 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
4 |
1 3
|
rabeqbidva |
⊢ ( 𝜑 → { 𝑥 ∈ 𝐴 ∣ 𝜓 } = { 𝑥 ∈ 𝐵 ∣ 𝜒 } ) |