Metamath Proof Explorer
Description: Membership in a class abstraction, using implicit substitution.
(Contributed by NM, 13-Sep-1995)
|
|
Ref |
Expression |
|
Hypotheses |
elab2g.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
|
|
elab2g.2 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
|
Assertion |
elab2g |
⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
elab2g.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
| 2 |
|
elab2g.2 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
| 3 |
2
|
eleq2i |
⊢ ( 𝐴 ∈ 𝐵 ↔ 𝐴 ∈ { 𝑥 ∣ 𝜑 } ) |
| 4 |
1
|
elabg |
⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) |
| 5 |
3 4
|
bitrid |
⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) ) |