Metamath Proof Explorer
Description: A chained subclass and equality deduction. (Contributed by NM, 25-Apr-2004)
|
|
Ref |
Expression |
|
Hypotheses |
sseqtrdi.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
|
|
sseqtrdi.2 |
⊢ 𝐵 = 𝐶 |
|
Assertion |
sseqtrdi |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sseqtrdi.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
| 2 |
|
sseqtrdi.2 |
⊢ 𝐵 = 𝐶 |
| 3 |
2
|
sseq2i |
⊢ ( 𝐴 ⊆ 𝐵 ↔ 𝐴 ⊆ 𝐶 ) |
| 4 |
1 3
|
sylib |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 ) |