Metamath Proof Explorer


Theorem fipreima

Description: Given a finite subset A of the range of a function, there exists a finite subset of the domain whose image is A . (Contributed by Jeff Madsen, 2-Sep-2009) (Revised by Stefan O'Rear, 22-Feb-2015)

Ref Expression
Assertion fipreima FFnBAranFAFinc𝒫BFinFc=A

Proof

Step Hyp Ref Expression
1 simp3 FFnBAranFAFinAFin
2 dfss3 AranFxAxranF
3 fvelrnb FFnBxranFyBFy=x
4 3 ralbidv FFnBxAxranFxAyBFy=x
5 2 4 bitrid FFnBAranFxAyBFy=x
6 5 biimpa FFnBAranFxAyBFy=x
7 6 3adant3 FFnBAranFAFinxAyBFy=x
8 fveqeq2 y=fxFy=xFfx=x
9 8 ac6sfi AFinxAyBFy=xff:ABxAFfx=x
10 1 7 9 syl2anc FFnBAranFAFinff:ABxAFfx=x
11 fimass f:ABfAB
12 vex fV
13 12 imaex fAV
14 13 elpw fA𝒫BfAB
15 11 14 sylibr f:ABfA𝒫B
16 15 ad2antrl FFnBAranFAFinf:ABxAFfx=xfA𝒫B
17 ffun f:ABFunf
18 17 ad2antrl FFnBAranFAFinf:ABxAFfx=xFunf
19 simpl3 FFnBAranFAFinf:ABxAFfx=xAFin
20 imafi FunfAFinfAFin
21 18 19 20 syl2anc FFnBAranFAFinf:ABxAFfx=xfAFin
22 16 21 elind FFnBAranFAFinf:ABxAFfx=xfA𝒫BFin
23 fvco3 f:ABxAFfx=Ffx
24 fvresi xAIAx=x
25 24 adantl f:ABxAIAx=x
26 23 25 eqeq12d f:ABxAFfx=IAxFfx=x
27 26 ralbidva f:ABxAFfx=IAxxAFfx=x
28 27 biimprd f:ABxAFfx=xxAFfx=IAx
29 28 adantl FFnBAranFAFinf:ABxAFfx=xxAFfx=IAx
30 29 impr FFnBAranFAFinf:ABxAFfx=xxAFfx=IAx
31 simpl1 FFnBAranFAFinf:ABxAFfx=xFFnB
32 ffn f:ABfFnA
33 32 ad2antrl FFnBAranFAFinf:ABxAFfx=xfFnA
34 frn f:ABranfB
35 34 ad2antrl FFnBAranFAFinf:ABxAFfx=xranfB
36 fnco FFnBfFnAranfBFfFnA
37 31 33 35 36 syl3anc FFnBAranFAFinf:ABxAFfx=xFfFnA
38 fnresi IAFnA
39 eqfnfv FfFnAIAFnAFf=IAxAFfx=IAx
40 37 38 39 sylancl FFnBAranFAFinf:ABxAFfx=xFf=IAxAFfx=IAx
41 30 40 mpbird FFnBAranFAFinf:ABxAFfx=xFf=IA
42 41 imaeq1d FFnBAranFAFinf:ABxAFfx=xFfA=IAA
43 imaco FfA=FfA
44 ssid AA
45 resiima AAIAA=A
46 44 45 ax-mp IAA=A
47 42 43 46 3eqtr3g FFnBAranFAFinf:ABxAFfx=xFfA=A
48 imaeq2 c=fAFc=FfA
49 48 eqeq1d c=fAFc=AFfA=A
50 49 rspcev fA𝒫BFinFfA=Ac𝒫BFinFc=A
51 22 47 50 syl2anc FFnBAranFAFinf:ABxAFfx=xc𝒫BFinFc=A
52 10 51 exlimddv FFnBAranFAFinc𝒫BFinFc=A