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Metamath Proof Explorer
Description: Equality of a class variable and a class abstraction (deduction form of
eqabb ). (Contributed by NM, 16-Nov-1995)
|
|
Ref |
Expression |
|
Hypothesis |
eqabrd.1 |
⊢ ( 𝜑 → 𝐴 = { 𝑥 ∣ 𝜓 } ) |
|
Assertion |
eqabrd |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝜓 ) ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
eqabrd.1 |
⊢ ( 𝜑 → 𝐴 = { 𝑥 ∣ 𝜓 } ) |
| 2 |
1
|
eleq2d |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝑥 ∈ { 𝑥 ∣ 𝜓 } ) ) |
| 3 |
|
abid |
⊢ ( 𝑥 ∈ { 𝑥 ∣ 𝜓 } ↔ 𝜓 ) |
| 4 |
2 3
|
bitrdi |
⊢ ( 𝜑 → ( 𝑥 ∈ 𝐴 ↔ 𝜓 ) ) |