Metamath Proof Explorer
Description: Equality deduction for intersection of two classes. (Contributed by NM, 10-Apr-1994)
|
|
Ref |
Expression |
|
Hypothesis |
ineq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
|
Assertion |
ineq2d |
⊢ ( 𝜑 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
ineq1d.1 |
⊢ ( 𝜑 → 𝐴 = 𝐵 ) |
2 |
|
ineq2 |
⊢ ( 𝐴 = 𝐵 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) ) |
3 |
1 2
|
syl |
⊢ ( 𝜑 → ( 𝐶 ∩ 𝐴 ) = ( 𝐶 ∩ 𝐵 ) ) |