Metamath Proof Explorer


Theorem imainrect

Description: Image by a restricted and corestricted binary relation (intersection of a binary relation with a Cartesian product). (Contributed by Stefan O'Rear, 19-Feb-2015)

Ref Expression
Assertion imainrect ⊢ G ∩ A × B Y = G Y ∩ A ∩ B

Proof

Step Hyp Ref Expression
1 df-res ⊢ G ∩ A × B ↾ Y = G ∩ A × B ∩ Y × V
2 1 rneqi ⊢ ran ⁡ G ∩ A × B ↾ Y = ran ⁡ G ∩ A × B ∩ Y × V
3 df-ima ⊢ G ∩ A × B Y = ran ⁡ G ∩ A × B ↾ Y
4 df-ima ⊢ G Y ∩ A = ran ⁡ G ↾ Y ∩ A
5 df-res ⊢ G ↾ Y ∩ A = G ∩ Y ∩ A × V
6 5 rneqi ⊢ ran ⁡ G ↾ Y ∩ A = ran ⁡ G ∩ Y ∩ A × V
7 4 6 eqtri ⊢ G Y ∩ A = ran ⁡ G ∩ Y ∩ A × V
8 7 ineq1i ⊢ G Y ∩ A ∩ B = ran ⁡ G ∩ Y ∩ A × V ∩ B
9 cnvin ⊢ G ∩ Y ∩ A × V ∩ V × B -1 = G ∩ Y ∩ A × V -1 ∩ V × B -1
10 inxp ⊢ A × V ∩ V × B = A ∩ V × V ∩ B
11 inv1 ⊢ A ∩ V = A
12 incom ⊢ V ∩ B = B ∩ V
13 inv1 ⊢ B ∩ V = B
14 12 13 eqtri ⊢ V ∩ B = B
15 11 14 xpeq12i ⊢ A ∩ V × V ∩ B = A × B
16 10 15 eqtr2i ⊢ A × B = A × V ∩ V × B
17 16 ineq2i ⊢ G ∩ Y × V ∩ A × B = G ∩ Y × V ∩ A × V ∩ V × B
18 in32 ⊢ G ∩ A × B ∩ Y × V = G ∩ Y × V ∩ A × B
19 xpindir ⊢ Y ∩ A × V = Y × V ∩ A × V
20 19 ineq2i ⊢ G ∩ Y ∩ A × V = G ∩ Y × V ∩ A × V
21 inass ⊢ G ∩ Y × V ∩ A × V = G ∩ Y × V ∩ A × V
22 20 21 eqtr4i ⊢ G ∩ Y ∩ A × V = G ∩ Y × V ∩ A × V
23 22 ineq1i ⊢ G ∩ Y ∩ A × V ∩ V × B = G ∩ Y × V ∩ A × V ∩ V × B
24 inass ⊢ G ∩ Y × V ∩ A × V ∩ V × B = G ∩ Y × V ∩ A × V ∩ V × B
25 23 24 eqtri ⊢ G ∩ Y ∩ A × V ∩ V × B = G ∩ Y × V ∩ A × V ∩ V × B
26 17 18 25 3eqtr4i ⊢ G ∩ A × B ∩ Y × V = G ∩ Y ∩ A × V ∩ V × B
27 26 cnveqi ⊢ G ∩ A × B ∩ Y × V -1 = G ∩ Y ∩ A × V ∩ V × B -1
28 df-res ⊢ G ∩ Y ∩ A × V -1 ↾ B = G ∩ Y ∩ A × V -1 ∩ B × V
29 cnvxp ⊢ V × B -1 = B × V
30 29 ineq2i ⊢ G ∩ Y ∩ A × V -1 ∩ V × B -1 = G ∩ Y ∩ A × V -1 ∩ B × V
31 28 30 eqtr4i ⊢ G ∩ Y ∩ A × V -1 ↾ B = G ∩ Y ∩ A × V -1 ∩ V × B -1
32 9 27 31 3eqtr4ri ⊢ G ∩ Y ∩ A × V -1 ↾ B = G ∩ A × B ∩ Y × V -1
33 32 dmeqi ⊢ dom ⁡ G ∩ Y ∩ A × V -1 ↾ B = dom ⁡ G ∩ A × B ∩ Y × V -1
34 incom ⊢ B ∩ dom ⁡ G ∩ Y ∩ A × V -1 = dom ⁡ G ∩ Y ∩ A × V -1 ∩ B
35 dmres ⊢ dom ⁡ G ∩ Y ∩ A × V -1 ↾ B = B ∩ dom ⁡ G ∩ Y ∩ A × V -1
36 df-rn ⊢ ran ⁡ G ∩ Y ∩ A × V = dom ⁡ G ∩ Y ∩ A × V -1
37 36 ineq1i ⊢ ran ⁡ G ∩ Y ∩ A × V ∩ B = dom ⁡ G ∩ Y ∩ A × V -1 ∩ B
38 34 35 37 3eqtr4ri ⊢ ran ⁡ G ∩ Y ∩ A × V ∩ B = dom ⁡ G ∩ Y ∩ A × V -1 ↾ B
39 df-rn ⊢ ran ⁡ G ∩ A × B ∩ Y × V = dom ⁡ G ∩ A × B ∩ Y × V -1
40 33 38 39 3eqtr4ri ⊢ ran ⁡ G ∩ A × B ∩ Y × V = ran ⁡ G ∩ Y ∩ A × V ∩ B
41 8 40 eqtr4i ⊢ G Y ∩ A ∩ B = ran ⁡ G ∩ A × B ∩ Y × V
42 2 3 41 3eqtr4i ⊢ G ∩ A × B Y = G Y ∩ A ∩ B