Metamath Proof Explorer
Description: Membership inference from subclass relationship. (Contributed by NM, 14-Dec-2004)
|
|
Ref |
Expression |
|
Hypotheses |
sseld.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
|
|
sseldd.2 |
⊢ ( 𝜑 → 𝐶 ∈ 𝐴 ) |
|
Assertion |
sseldd |
⊢ ( 𝜑 → 𝐶 ∈ 𝐵 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sseld.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
| 2 |
|
sseldd.2 |
⊢ ( 𝜑 → 𝐶 ∈ 𝐴 ) |
| 3 |
1
|
sseld |
⊢ ( 𝜑 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) ) |
| 4 |
2 3
|
mpd |
⊢ ( 𝜑 → 𝐶 ∈ 𝐵 ) |