Metamath Proof Explorer
Description: Membership in a class abstraction, using implicit substitution.
Deduction version of elab . (Contributed by Gino Giotto, 12-Oct-2024)
|
|
Ref |
Expression |
|
Hypotheses |
elabd3.ex |
⊢ ( 𝜑 → 𝐴 ∈ 𝑉 ) |
|
|
elabd3.is |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
|
Assertion |
elabd3 |
⊢ ( 𝜑 → ( 𝐴 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜒 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
elabd3.ex |
⊢ ( 𝜑 → 𝐴 ∈ 𝑉 ) |
2 |
|
elabd3.is |
⊢ ( ( 𝜑 ∧ 𝑥 = 𝐴 ) → ( 𝜓 ↔ 𝜒 ) ) |
3 |
|
eqidd |
⊢ ( 𝜑 → { 𝑥 ∣ 𝜓 } = { 𝑥 ∣ 𝜓 } ) |
4 |
1 3 2
|
elabd2 |
⊢ ( 𝜑 → ( 𝐴 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜒 ) ) |