Metamath Proof Explorer
Description: Membership in a class abstraction, using implicit substitution.
(Contributed by NM, 13-Sep-1995)
|
|
Ref |
Expression |
|
Hypotheses |
elab2.1 |
⊢ 𝐴 ∈ V |
|
|
elab2.2 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
|
|
elab2.3 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
|
Assertion |
elab2 |
⊢ ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
elab2.1 |
⊢ 𝐴 ∈ V |
2 |
|
elab2.2 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
3 |
|
elab2.3 |
⊢ 𝐵 = { 𝑥 ∣ 𝜑 } |
4 |
2 3
|
elab2g |
⊢ ( 𝐴 ∈ V → ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) ) |
5 |
1 4
|
ax-mp |
⊢ ( 𝐴 ∈ 𝐵 ↔ 𝜓 ) |