Metamath Proof Explorer


Theorem brimg

Description: Binary relation form of the Img function. (Contributed by Scott Fenton, 12-Apr-2014) (Revised by Mario Carneiro, 19-Apr-2014) (Proof shortened by Peter Mazsa, 2-Oct-2022)

Ref Expression
Hypotheses brimg.1 AV
brimg.2 BV
brimg.3 CV
Assertion brimg AB𝖨𝗆𝗀CC=AB

Proof

Step Hyp Ref Expression
1 brimg.1 AV
2 brimg.2 BV
3 brimg.3 CV
4 df-img 𝖨𝗆𝗀=𝖨𝗆𝖺𝗀𝖾2nd1st1stV×V𝖢𝖺𝗋𝗍
5 4 breqi AB𝖨𝗆𝗀CAB𝖨𝗆𝖺𝗀𝖾2nd1st1stV×V𝖢𝖺𝗋𝗍C
6 opex ABV
7 6 3 brco AB𝖨𝗆𝖺𝗀𝖾2nd1st1stV×V𝖢𝖺𝗋𝗍CaAB𝖢𝖺𝗋𝗍aa𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
8 vex aV
9 1 2 8 brcart AB𝖢𝖺𝗋𝗍aa=A×B
10 9 anbi1i AB𝖢𝖺𝗋𝗍aa𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCa=A×Ba𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
11 10 exbii aAB𝖢𝖺𝗋𝗍aa𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCaa=A×Ba𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
12 1 2 xpex A×BV
13 breq1 a=A×Ba𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCA×B𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
14 12 13 ceqsexv aa=A×Ba𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCA×B𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
15 7 11 14 3bitri AB𝖨𝗆𝖺𝗀𝖾2nd1st1stV×V𝖢𝖺𝗋𝗍CA×B𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VC
16 12 3 brimage A×B𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCC=2nd1st1stV×VA×B
17 19.42v abBaAab2nd1st1stV×VxbBaaAab2nd1st1stV×Vx
18 anass p=abaAbBp2nd1st1stV×Vxp=abaAbBp2nd1st1stV×Vx
19 an21 aAbBp2nd1st1stV×VxbBaAp2nd1st1stV×Vx
20 19 anbi2i p=abaAbBp2nd1st1stV×Vxp=abbBaAp2nd1st1stV×Vx
21 18 20 bitri p=abaAbBp2nd1st1stV×Vxp=abbBaAp2nd1st1stV×Vx
22 21 2exbii pap=abaAbBp2nd1st1stV×Vxpap=abbBaAp2nd1st1stV×Vx
23 excom pap=abbBaAp2nd1st1stV×Vxapp=abbBaAp2nd1st1stV×Vx
24 opex abV
25 breq1 p=abp2nd1st1stV×Vxab2nd1st1stV×Vx
26 25 anbi2d p=abaAp2nd1st1stV×VxaAab2nd1st1stV×Vx
27 26 anbi2d p=abbBaAp2nd1st1stV×VxbBaAab2nd1st1stV×Vx
28 24 27 ceqsexv pp=abbBaAp2nd1st1stV×VxbBaAab2nd1st1stV×Vx
29 28 exbii app=abbBaAp2nd1st1stV×VxabBaAab2nd1st1stV×Vx
30 22 23 29 3bitri pap=abaAbBp2nd1st1stV×VxabBaAab2nd1st1stV×Vx
31 df-br bAxbxA
32 risset bxAaAa=bx
33 vex bV
34 33 brresi a1stV×VbaV×Va1stb
35 df-br a1stV×Vbab1stV×V
36 34 35 bitr3i aV×Va1stbab1stV×V
37 36 anbi1i aV×Va1stbab2nd1stxab1stV×Vab2nd1stx
38 elvv aV×Vpqa=pq
39 38 anbi1i aV×Va1stbab2nd1stxpqa=pqa1stbab2nd1stx
40 anass aV×Va1stbab2nd1stxaV×Va1stbab2nd1stx
41 ancom a=pqp=bq=xp=bq=xa=pq
42 breq1 a=pqa1stbpq1stb
43 opeq1 a=pqab=pqb
44 43 breq1d a=pqab2nd1stxpqb2nd1stx
45 42 44 anbi12d a=pqa1stbab2nd1stxpq1stbpqb2nd1stx
46 vex pV
47 vex qV
48 46 47 br1steq pq1stbb=p
49 equcom b=pp=b
50 48 49 bitri pq1stbp=b
51 opex pqbV
52 vex xV
53 51 52 brco pqb2nd1stxapqb1staa2ndx
54 opex pqV
55 54 33 br1steq pqb1staa=pq
56 55 anbi1i pqb1staa2ndxa=pqa2ndx
57 56 exbii apqb1staa2ndxaa=pqa2ndx
58 53 57 bitri pqb2nd1stxaa=pqa2ndx
59 breq1 a=pqa2ndxpq2ndx
60 54 59 ceqsexv aa=pqa2ndxpq2ndx
61 46 47 br2ndeq pq2ndxx=q
62 60 61 bitri aa=pqa2ndxx=q
63 equcom x=qq=x
64 58 62 63 3bitri pqb2nd1stxq=x
65 50 64 anbi12i pq1stbpqb2nd1stxp=bq=x
66 45 65 bitrdi a=pqa1stbab2nd1stxp=bq=x
67 66 pm5.32i a=pqa1stbab2nd1stxa=pqp=bq=x
68 df-3an p=bq=xa=pqp=bq=xa=pq
69 41 67 68 3bitr4i a=pqa1stbab2nd1stxp=bq=xa=pq
70 69 2exbii pqa=pqa1stbab2nd1stxpqp=bq=xa=pq
71 19.41vv pqa=pqa1stbab2nd1stxpqa=pqa1stbab2nd1stx
72 opeq1 p=bpq=bq
73 72 eqeq2d p=ba=pqa=bq
74 opeq2 q=xbq=bx
75 74 eqeq2d q=xa=bqa=bx
76 33 52 73 75 ceqsex2v pqp=bq=xa=pqa=bx
77 70 71 76 3bitr3ri a=bxpqa=pqa1stbab2nd1stx
78 39 40 77 3bitr4ri a=bxaV×Va1stbab2nd1stx
79 52 brresi ab2nd1st1stV×Vxab1stV×Vab2nd1stx
80 37 78 79 3bitr4i a=bxab2nd1st1stV×Vx
81 80 rexbii aAa=bxaAab2nd1st1stV×Vx
82 32 81 bitri bxAaAab2nd1st1stV×Vx
83 df-rex aAab2nd1st1stV×VxaaAab2nd1st1stV×Vx
84 31 82 83 3bitri bAxaaAab2nd1st1stV×Vx
85 84 anbi2i bBbAxbBaaAab2nd1st1stV×Vx
86 17 30 85 3bitr4ri bBbAxpap=abaAbBp2nd1st1stV×Vx
87 86 exbii bbBbAxbpap=abaAbBp2nd1st1stV×Vx
88 52 elima2 xABbbBbAx
89 52 elima2 x2nd1st1stV×VA×BppA×Bp2nd1st1stV×Vx
90 elxp pA×Babp=abaAbB
91 90 anbi1i pA×Bp2nd1st1stV×Vxabp=abaAbBp2nd1st1stV×Vx
92 19.41vv abp=abaAbBp2nd1st1stV×Vxabp=abaAbBp2nd1st1stV×Vx
93 91 92 bitr4i pA×Bp2nd1st1stV×Vxabp=abaAbBp2nd1st1stV×Vx
94 93 exbii ppA×Bp2nd1st1stV×Vxpabp=abaAbBp2nd1st1stV×Vx
95 exrot3 pabp=abaAbBp2nd1st1stV×Vxabpp=abaAbBp2nd1st1stV×Vx
96 exrot3 abpp=abaAbBp2nd1st1stV×Vxbpap=abaAbBp2nd1st1stV×Vx
97 95 96 bitri pabp=abaAbBp2nd1st1stV×Vxbpap=abaAbBp2nd1st1stV×Vx
98 89 94 97 3bitri x2nd1st1stV×VA×Bbpap=abaAbBp2nd1st1stV×Vx
99 87 88 98 3bitr4ri x2nd1st1stV×VA×BxAB
100 99 eqriv 2nd1st1stV×VA×B=AB
101 100 eqeq2i C=2nd1st1stV×VA×BC=AB
102 16 101 bitri A×B𝖨𝗆𝖺𝗀𝖾2nd1st1stV×VCC=AB
103 5 15 102 3bitri AB𝖨𝗆𝗀CC=AB