Metamath Proof Explorer


Theorem btwnconn1lem7

Description: Lemma for btwnconn1 . Under our assumptions, C and d are distinct. (Contributed by Scott Fenton, 8-Oct-2013)

Ref Expression
Assertion btwnconn1lem7 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCd

Proof

Step Hyp Ref Expression
1 simp1l3 ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBCc
2 1 adantr ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCc
3 simp2rr ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBCdCgrCD
4 3 adantr ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCdCgrCD
5 simp2lr ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBDcCgrCD
6 5 adantr ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdDcCgrCD
7 2 4 6 3jca ABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCcCdCgrCDDcCgrCD
8 simp11 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NN
9 simp21 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NC𝔼N
10 simp22 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼ND𝔼N
11 simp23 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼Nc𝔼N
12 simp31 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼Nd𝔼N
13 simpr1 NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDCc
14 opeq2 C=dCC=Cd
15 14 breq1d C=dCCCgrCDCdCgrCD
16 15 3anbi2d C=dCcCCCgrCDDcCgrCDCcCdCgrCDDcCgrCD
17 16 biimparc CcCdCgrCDDcCgrCDC=dCcCCCgrCDDcCgrCD
18 simp2 CcCCCgrCDDcCgrCDCCCgrCD
19 simp1 NC𝔼ND𝔼Nc𝔼Nd𝔼NN
20 simp2l NC𝔼ND𝔼Nc𝔼Nd𝔼NC𝔼N
21 simp2r NC𝔼ND𝔼Nc𝔼Nd𝔼ND𝔼N
22 cgrid2 NC𝔼NC𝔼ND𝔼NCCCgrCDC=D
23 19 20 20 21 22 syl13anc NC𝔼ND𝔼Nc𝔼Nd𝔼NCCCgrCDC=D
24 18 23 syl5 NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCCCgrCDDcCgrCDC=D
25 24 imp NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCCCgrCDDcCgrCDC=D
26 opeq1 C=DCc=Dc
27 opeq2 C=DCC=CD
28 26 27 breq12d C=DCcCgrCCDcCgrCD
29 28 biimparc DcCgrCDC=DCcCgrCC
30 simp3l NC𝔼ND𝔼Nc𝔼Nd𝔼Nc𝔼N
31 axcgrid NC𝔼Nc𝔼NC𝔼NCcCgrCCC=c
32 19 20 30 20 31 syl13anc NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCgrCCC=c
33 29 32 syl5 NC𝔼ND𝔼Nc𝔼Nd𝔼NDcCgrCDC=DC=c
34 33 expdimp NC𝔼ND𝔼Nc𝔼Nd𝔼NDcCgrCDC=DC=c
35 34 3ad2antr3 NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCCCgrCDDcCgrCDC=DC=c
36 25 35 mpd NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCCCgrCDDcCgrCDC=c
37 36 ex NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCCCgrCDDcCgrCDC=c
38 17 37 syl5 NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDC=dC=c
39 38 expdimp NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDC=dC=c
40 39 necon3d NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDCcCd
41 13 40 mpd NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDCd
42 41 ex NC𝔼ND𝔼Nc𝔼Nd𝔼NCcCdCgrCDDcCgrCDCd
43 8 9 10 11 12 42 syl122anc NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NCcCdCgrCDDcCgrCDCd
44 7 43 syl5 NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCd
45 44 imp NA𝔼NB𝔼NC𝔼ND𝔼Nc𝔼Nd𝔼Nb𝔼NE𝔼NABBCCcBBtwnACBBtwnADDBtwnAcDcCgrCDCBtwnAdCdCgrCDcBtwnAbcbCgrCBdBtwnAbdbCgrDBEBtwnCcEBtwnDdCd