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
|
biimpi |
⊢ ( ran 𝐹 ⊆ 𝐵 → ( 𝐵 ∩ ran 𝐹 ) = ran 𝐹 ) |
4 |
3
|
adantl |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( 𝐵 ∩ ran 𝐹 ) = ran 𝐹 ) |
5 |
4
|
ineq2d |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( 𝐴 ∩ ( 𝐵 ∩ ran 𝐹 ) ) = ( 𝐴 ∩ ran 𝐹 ) ) |
6 |
1 5
|
eqtrid |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) = ( 𝐴 ∩ ran 𝐹 ) ) |
7 |
6
|
imaeq2d |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
8 |
|
fimacnvinrn |
⊢ ( Fun 𝐹 → ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) = ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) ) |
9 |
8
|
adantr |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) = ( ◡ 𝐹 “ ( ( 𝐴 ∩ 𝐵 ) ∩ ran 𝐹 ) ) ) |
10 |
|
fimacnvinrn |
⊢ ( Fun 𝐹 → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
11 |
10
|
adantr |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ ran 𝐹 ) ) ) |
12 |
7 9 11
|
3eqtr4rd |
⊢ ( ( Fun 𝐹 ∧ ran 𝐹 ⊆ 𝐵 ) → ( ◡ 𝐹 “ 𝐴 ) = ( ◡ 𝐹 “ ( 𝐴 ∩ 𝐵 ) ) ) |