Metamath Proof Explorer
Description: Membership deduction from subclass relationship. (Contributed by NM, 26-Jun-2014)
|
|
Ref |
Expression |
|
Hypothesis |
sseld.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
|
Assertion |
sselda |
⊢ ( ( 𝜑 ∧ 𝐶 ∈ 𝐴 ) → 𝐶 ∈ 𝐵 ) |
Proof
| Step |
Hyp |
Ref |
Expression |
| 1 |
|
sseld.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
| 2 |
1
|
sseld |
⊢ ( 𝜑 → ( 𝐶 ∈ 𝐴 → 𝐶 ∈ 𝐵 ) ) |
| 3 |
2
|
imp |
⊢ ( ( 𝜑 ∧ 𝐶 ∈ 𝐴 ) → 𝐶 ∈ 𝐵 ) |