Metamath Proof Explorer


Theorem bnj1321

Description: Technical lemma for bnj60 . This lemma may no longer be used or have become an indirect lemma of the theorem in question (i.e. a lemma of a lemma... of the theorem). (Contributed by Jonathan Ben-Naim, 3-Jun-2011) (New usage is discouraged.)

Ref Expression
Hypotheses bnj1321.1 B=d|dAxdpredxARd
bnj1321.2 Y=xfpredxAR
bnj1321.3 C=f|dBfFndxdfx=GY
bnj1321.4 τfCdomf=xtrClxAR
Assertion bnj1321 RFrSeAfτ∃!fτ

Proof

Step Hyp Ref Expression
1 bnj1321.1 B=d|dAxdpredxARd
2 bnj1321.2 Y=xfpredxAR
3 bnj1321.3 C=f|dBfFndxdfx=GY
4 bnj1321.4 τfCdomf=xtrClxAR
5 simpr RFrSeAfτfτ
6 simp1 RFrSeAτgfτRFrSeA
7 4 simplbi τfC
8 7 3ad2ant2 RFrSeAτgfτfC
9 nfab1 _ff|dBfFndxdfx=GY
10 3 9 nfcxfr _fC
11 10 nfcri fgC
12 nfv fdomg=xtrClxAR
13 11 12 nfan fgCdomg=xtrClxAR
14 eleq1w f=gfCgC
15 dmeq f=gdomf=domg
16 15 eqeq1d f=gdomf=xtrClxARdomg=xtrClxAR
17 14 16 anbi12d f=gfCdomf=xtrClxARgCdomg=xtrClxAR
18 4 17 bitrid f=gτgCdomg=xtrClxAR
19 13 18 sbiev gfτgCdomg=xtrClxAR
20 19 simplbi gfτgC
21 20 3ad2ant3 RFrSeAτgfτgC
22 eqid domfdomg=domfdomg
23 1 2 3 22 bnj1326 RFrSeAfCgCfdomfdomg=gdomfdomg
24 6 8 21 23 syl3anc RFrSeAτgfτfdomfdomg=gdomfdomg
25 4 simprbi τdomf=xtrClxAR
26 25 3ad2ant2 RFrSeAτgfτdomf=xtrClxAR
27 19 simprbi gfτdomg=xtrClxAR
28 27 3ad2ant3 RFrSeAτgfτdomg=xtrClxAR
29 26 28 eqtr4d RFrSeAτgfτdomf=domg
30 bnj1322 domf=domgdomfdomg=domf
31 30 reseq2d domf=domgfdomfdomg=fdomf
32 29 31 syl RFrSeAτgfτfdomfdomg=fdomf
33 releq z=fRelzRelf
34 1 2 3 bnj66 zCRelz
35 33 34 vtoclga fCRelf
36 resdm Relffdomf=f
37 8 35 36 3syl RFrSeAτgfτfdomf=f
38 32 37 eqtrd RFrSeAτgfτfdomfdomg=f
39 eqeq2 domf=domgdomfdomg=domfdomfdomg=domg
40 30 39 mpbid domf=domgdomfdomg=domg
41 40 reseq2d domf=domggdomfdomg=gdomg
42 29 41 syl RFrSeAτgfτgdomfdomg=gdomg
43 1 2 3 bnj66 gCRelg
44 resdm Relggdomg=g
45 21 43 44 3syl RFrSeAτgfτgdomg=g
46 42 45 eqtrd RFrSeAτgfτgdomfdomg=g
47 24 38 46 3eqtr3d RFrSeAτgfτf=g
48 47 3expib RFrSeAτgfτf=g
49 48 alrimivv RFrSeAfgτgfτf=g
50 49 adantr RFrSeAfτfgτgfτf=g
51 nfv gτ
52 51 eu2 ∃!fτfτfgτgfτf=g
53 5 50 52 sylanbrc RFrSeAfτ∃!fτ