Metamath Proof Explorer
Description: Membership in a class abstraction, using implicit substitution.
(Contributed by NM, 1-Aug-1994) (Revised by Mario Carneiro, 12-Oct-2016)
|
|
Ref |
Expression |
|
Hypotheses |
elabf.1 |
⊢ Ⅎ 𝑥 𝜓 |
|
|
elabf.2 |
⊢ 𝐴 ∈ V |
|
|
elabf.3 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
|
Assertion |
elabf |
⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
elabf.1 |
⊢ Ⅎ 𝑥 𝜓 |
2 |
|
elabf.2 |
⊢ 𝐴 ∈ V |
3 |
|
elabf.3 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
4 |
|
nfcv |
⊢ Ⅎ 𝑥 𝐴 |
5 |
4 1 3
|
elabgf |
⊢ ( 𝐴 ∈ V → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) |
6 |
2 5
|
ax-mp |
⊢ ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) |