Metamath Proof Explorer
Description: Subclass theorem for disjoint elementhood, deduction version.
(Contributed by Peter Mazsa, 28-Sep-2021)
|
|
Ref |
Expression |
|
Hypothesis |
eldisjssd.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
|
Assertion |
eldisjssd |
⊢ ( 𝜑 → ( ElDisj 𝐵 → ElDisj 𝐴 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
eldisjssd.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
2 |
|
eldisjss |
⊢ ( 𝐴 ⊆ 𝐵 → ( ElDisj 𝐵 → ElDisj 𝐴 ) ) |
3 |
1 2
|
syl |
⊢ ( 𝜑 → ( ElDisj 𝐵 → ElDisj 𝐴 ) ) |