Metamath Proof Explorer
Description: An equality inference for the subclass relationship. (Contributed by NM, 30-Aug-1993)
|
|
Ref |
Expression |
|
Hypothesis |
sseq1i.1 |
⊢ 𝐴 = 𝐵 |
|
Assertion |
sseq2i |
⊢ ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sseq1i.1 |
⊢ 𝐴 = 𝐵 |
| 2 |
|
sseq2 |
⊢ ( 𝐴 = 𝐵 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) ) |
| 3 |
1 2
|
ax-mp |
⊢ ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) |