Metamath Proof Explorer


Theorem sbthlem6

Description: Lemma for sbth . (Contributed by NM, 27-Mar-1998)

Ref Expression
Hypotheses sbthlem.1 AV
sbthlem.2 D=x|xAgBfxAx
sbthlem.3 H=fDg-1AD
Assertion sbthlem6 ranfBdomg=BrangAFung-1ranH=B

Proof

Step Hyp Ref Expression
1 sbthlem.1 AV
2 sbthlem.2 D=x|xAgBfxAx
3 sbthlem.3 H=fDg-1AD
4 rnun ranfDg-1AD=ranfDrang-1AD
5 3 rneqi ranH=ranfDg-1AD
6 df-ima fD=ranfD
7 6 uneq1i fDrang-1AD=ranfDrang-1AD
8 4 5 7 3eqtr4i ranH=fDrang-1AD
9 1 2 sbthlem4 domg=BrangAFung-1g-1AD=BfD
10 df-ima g-1AD=rang-1AD
11 9 10 eqtr3di domg=BrangAFung-1BfD=rang-1AD
12 11 uneq2d domg=BrangAFung-1fDBfD=fDrang-1AD
13 8 12 eqtr4id domg=BrangAFung-1ranH=fDBfD
14 imassrn fDranf
15 sstr2 fDranfranfBfDB
16 14 15 ax-mp ranfBfDB
17 undif fDBfDBfD=B
18 16 17 sylib ranfBfDBfD=B
19 13 18 sylan9eqr ranfBdomg=BrangAFung-1ranH=B