Metamath Proof Explorer
Description: Deduction version of elrab , just like elrabd , but backwards
direction. (Contributed by Thierry Arnoux, 15-Jan-2026)
|
|
Ref |
Expression |
|
Hypotheses |
elrabrd.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) |
|
|
elrabrd.2 |
⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |
|
Assertion |
elrabrd |
⊢ ( 𝜑 → 𝜒 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
elrabrd.1 |
⊢ ( 𝑥 = 𝐴 → ( 𝜓 ↔ 𝜒 ) ) |
| 2 |
|
elrabrd.2 |
⊢ ( 𝜑 → 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ) |
| 3 |
1
|
elrab |
⊢ ( 𝐴 ∈ { 𝑥 ∈ 𝐵 ∣ 𝜓 } ↔ ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) ) |
| 4 |
2 3
|
sylib |
⊢ ( 𝜑 → ( 𝐴 ∈ 𝐵 ∧ 𝜒 ) ) |
| 5 |
4
|
simprd |
⊢ ( 𝜑 → 𝜒 ) |