Metamath Proof Explorer


Theorem f1omptsnlem

Description: This is the core of the proof of f1omptsn , but to avoid the distinct variables on the definitions, we split this proof into two. (Contributed by ML, 15-Jul-2020)

Ref Expression
Hypotheses f1omptsn.f F=xAx
f1omptsn.r R=u|xAu=x
Assertion f1omptsnlem F:A1-1 ontoR

Proof

Step Hyp Ref Expression
1 f1omptsn.f F=xAx
2 f1omptsn.r R=u|xAu=x
3 eqid x=x
4 vsnex xV
5 eqsbc1 xV[˙x/u]˙u=xx=x
6 4 5 ax-mp [˙x/u]˙u=xx=x
7 3 6 mpbir [˙x/u]˙u=x
8 sbcel2 [˙x/u]˙xAxx/uA
9 csbconstg xVx/uA=A
10 4 9 ax-mp x/uA=A
11 10 eleq2i xx/uAxA
12 8 11 bitri [˙x/u]˙xAxA
13 2 eqabri uRxAu=x
14 df-rex xAu=xxxAu=x
15 13 14 sylbbr xxAu=xuR
16 15 19.23bi xAu=xuR
17 16 sbcth xV[˙x/u]˙xAu=xuR
18 4 17 ax-mp [˙x/u]˙xAu=xuR
19 sbcimg xV[˙x/u]˙xAu=xuR[˙x/u]˙xAu=x[˙x/u]˙uR
20 4 19 ax-mp [˙x/u]˙xAu=xuR[˙x/u]˙xAu=x[˙x/u]˙uR
21 18 20 mpbi [˙x/u]˙xAu=x[˙x/u]˙uR
22 sbcan [˙x/u]˙xAu=x[˙x/u]˙xA[˙x/u]˙u=x
23 sbcel1v [˙x/u]˙uRxR
24 21 22 23 3imtr3i [˙x/u]˙xA[˙x/u]˙u=xxR
25 12 24 sylanbr xA[˙x/u]˙u=xxR
26 7 25 mpan2 xAxR
27 1 26 fmpti F:AR
28 1 fvmpt2 xAxRFx=x
29 26 28 mpdan xAFx=x
30 sneq x=yx=y
31 30 1 4 fvmpt3i yAFy=y
32 29 31 eqeqan12d xAyAFx=Fyx=y
33 vex xV
34 sneqbg xVx=yx=y
35 33 34 ax-mp x=yx=y
36 32 35 bitrdi xAyAFx=Fyx=y
37 36 biimpd xAyAFx=Fyx=y
38 37 rgen2 xAyAFx=Fyx=y
39 dff13 F:A1-1RF:ARxAyAFx=Fyx=y
40 27 38 39 mpbir2an F:A1-1R
41 f1f1orn F:A1-1RF:A1-1 ontoranF
42 40 41 ax-mp F:A1-1 ontoranF
43 rnmptsn ranxAx=u|xAu=x
44 1 rneqi ranF=ranxAx
45 43 44 2 3eqtr4i ranF=R
46 f1oeq3 ranF=RF:A1-1 ontoranFF:A1-1 ontoR
47 45 46 ax-mp F:A1-1 ontoranFF:A1-1 ontoR
48 42 47 mpbi F:A1-1 ontoR