Metamath Proof Explorer
Description: Membership in a class abstraction, with a weaker antecedent than
elabg . (Contributed by NM, 29-Aug-2006)
|
|
Ref |
Expression |
|
Hypothesis |
elab3g.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
|
Assertion |
elab3g |
⊢ ( ( 𝜓 → 𝐴 ∈ 𝐵 ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
elab3g.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) ) |
2 |
|
nfcv |
⊢ Ⅎ 𝑥 𝐴 |
3 |
|
nfv |
⊢ Ⅎ 𝑥 𝜓 |
4 |
2 3 1
|
elab3gf |
⊢ ( ( 𝜓 → 𝐴 ∈ 𝐵 ) → ( 𝐴 ∈ { 𝑥 ∣ 𝜑 } ↔ 𝜓 ) ) |