Metamath Proof Explorer
Description: An equality inference for the subclass relationship. (Contributed by NM, 31-May-1999) (Proof shortened by Eric Schmidt, 26-Jan-2007)
|
|
Ref |
Expression |
|
Hypotheses |
sseq1i.1 |
⊢ 𝐴 = 𝐵 |
|
|
sseq12i.2 |
⊢ 𝐶 = 𝐷 |
|
Assertion |
sseq12i |
⊢ ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sseq1i.1 |
⊢ 𝐴 = 𝐵 |
2 |
|
sseq12i.2 |
⊢ 𝐶 = 𝐷 |
3 |
|
sseq12 |
⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 ) ) |
4 |
1 2 3
|
mp2an |
⊢ ( 𝐴 ⊆ 𝐶 ↔ 𝐵 ⊆ 𝐷 ) |