Metamath Proof Explorer
Description: Equality theorem for the collection operation. (Contributed by Rohan
Ridenour, 11-Aug-2023)
|
|
Ref |
Expression |
|
Assertion |
colleq1 |
⊢ ( 𝐹 = 𝐺 → ( 𝐹 Coll 𝐴 ) = ( 𝐺 Coll 𝐴 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
id |
⊢ ( 𝐹 = 𝐺 → 𝐹 = 𝐺 ) |
2 |
|
eqidd |
⊢ ( 𝐹 = 𝐺 → 𝐴 = 𝐴 ) |
3 |
1 2
|
colleq12d |
⊢ ( 𝐹 = 𝐺 → ( 𝐹 Coll 𝐴 ) = ( 𝐺 Coll 𝐴 ) ) |