Metamath Proof Explorer


Theorem sbthlem5

Description: Lemma for sbth . (Contributed by NM, 22-Mar-1998)

Ref Expression
Hypotheses sbthlem.1 AV
sbthlem.2 D=x|xAgBfxAx
sbthlem.3 H=fDg-1AD
Assertion sbthlem5 domf=ArangAdomH=A

Proof

Step Hyp Ref Expression
1 sbthlem.1 AV
2 sbthlem.2 D=x|xAgBfxAx
3 sbthlem.3 H=fDg-1AD
4 3 dmeqi domH=domfDg-1AD
5 dmun domfDg-1AD=domfDdomg-1AD
6 dmres domfD=Ddomf
7 dmres domg-1AD=ADdomg-1
8 df-rn rang=domg-1
9 8 eqcomi domg-1=rang
10 9 ineq2i ADdomg-1=ADrang
11 7 10 eqtri domg-1AD=ADrang
12 6 11 uneq12i domfDdomg-1AD=DdomfADrang
13 5 12 eqtri domfDg-1AD=DdomfADrang
14 4 13 eqtri domH=DdomfADrang
15 1 2 sbthlem1 DAgBfD
16 difss AgBfDA
17 15 16 sstri DA
18 sseq2 domf=ADdomfDA
19 17 18 mpbiri domf=ADdomf
20 dfss DdomfD=Ddomf
21 19 20 sylib domf=AD=Ddomf
22 21 uneq1d domf=ADAD=DdomfAD
23 1 2 sbthlem3 rangAgBfD=AD
24 imassrn gBfDrang
25 23 24 eqsstrrdi rangAADrang
26 dfss ADrangAD=ADrang
27 25 26 sylib rangAAD=ADrang
28 27 uneq2d rangADdomfAD=DdomfADrang
29 22 28 sylan9eq domf=ArangADAD=DdomfADrang
30 14 29 eqtr4id domf=ArangAdomH=DAD
31 undif DADAD=A
32 17 31 mpbi DAD=A
33 30 32 eqtrdi domf=ArangAdomH=A