| Step |
Hyp |
Ref |
Expression |
| 1 |
|
inass |
⊢ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) = ( 𝐴 ∩ ( 𝐵 ∩ ran 𝐹 ) ) |
| 2 |
|
sseqin2 |
⊢ ( ran 𝐹 ⊆ 𝐵 ↔ ( 𝐵 ∩ ran 𝐹 ) = ran 𝐹 ) |
| 3 |
2
|
bilani |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( 𝐵 ∩ ran 𝐹 ) = ran 𝐹 ) |
| 4 |
3
|
ineq2d |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( 𝐴 ∩ ( 𝐵 ∩ ran 𝐹 ) ) = ( 𝐴 ∩ ran 𝐹 ) ) |
| 5 |
1 4
|
eqtrid |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) = ( 𝐴 ∩ ran 𝐹 ) ) |
| 6 |
5
|
imaeq2d |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
| 7 |
|
fimacnvinrn |
⊢ ( Fun 𝐹 → ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) = ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) ) |
| 8 |
7
|
adantr |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) = ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) ) |
| 9 |
|
fimacnvinrn |
⊢ ( Fun 𝐹 → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
| 10 |
9
|
adantr |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
| 11 |
6 8 10
|
3eqtr4rd |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) ) |