| Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
| Mirrors > Home > MPE Home > Th. List > eqrdv | Structured version Visualization version GIF version | ||
| Description: Deduce equality of classes from equivalence of membership. (Contributed by NM, 17-Mar-1996.) |
| Ref | Expression |
|---|---|
| eqrdv.1 | ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) |
| Ref | Expression |
|---|---|
| eqrdv | ⊢ (𝜑 → 𝐴 = 𝐵) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | eqrdv.1 | . . 3 ⊢ (𝜑 → (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) | |
| 2 | 1 | alrimiv 1927 | . 2 ⊢ (𝜑 → ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) |
| 3 | dfcleq 2730 | . 2 ⊢ (𝐴 = 𝐵 ↔ ∀𝑥(𝑥 ∈ 𝐴 ↔ 𝑥 ∈ 𝐵)) | |
| 4 | 2, 3 | sylibr 234 | 1 ⊢ (𝜑 → 𝐴 = 𝐵) |
| Copyright terms: Public domain | W3C validator |