Metamath Proof Explorer
Description: Substitution of equality into a subclass relationship. (Contributed by NM, 25-Apr-2004)
|
|
Ref |
Expression |
|
Hypotheses |
sseqtrrd.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
|
|
sseqtrrd.2 |
⊢ ( 𝜑 → 𝐶 = 𝐵 ) |
|
Assertion |
sseqtrrd |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
sseqtrrd.1 |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐵 ) |
2 |
|
sseqtrrd.2 |
⊢ ( 𝜑 → 𝐶 = 𝐵 ) |
3 |
2
|
eqcomd |
⊢ ( 𝜑 → 𝐵 = 𝐶 ) |
4 |
1 3
|
sseqtrd |
⊢ ( 𝜑 → 𝐴 ⊆ 𝐶 ) |