Metamath Proof Explorer


Theorem pexmidlem8N

Description: Lemma for pexmidN . The contradiction of pexmidlem6N and pexmidlem7N shows that there can be no atom p that is not in X .+ ( ._|_X ) , which is therefore the whole atom space. (Contributed by NM, 3-Feb-2012) (New usage is discouraged.)

Ref Expression
Hypotheses pexmidALT.a A=AtomsK
pexmidALT.p +˙=+𝑃K
pexmidALT.o ˙=𝑃K
Assertion pexmidlem8N KHLXA˙˙X=XXX+˙˙X=A

Proof

Step Hyp Ref Expression
1 pexmidALT.a A=AtomsK
2 pexmidALT.p +˙=+𝑃K
3 pexmidALT.o ˙=𝑃K
4 nonconne ¬X=XXX
5 simpll KHLXA˙˙X=XXKHL
6 simplr KHLXA˙˙X=XXXA
7 1 3 polssatN KHLXA˙XA
8 7 adantr KHLXA˙˙X=XX˙XA
9 1 2 paddssat KHLXA˙XAX+˙˙XA
10 5 6 8 9 syl3anc KHLXA˙˙X=XXX+˙˙XA
11 df-pss X+˙˙XAX+˙˙XAX+˙˙XA
12 pssnel X+˙˙XAppA¬pX+˙˙X
13 11 12 sylbir X+˙˙XAX+˙˙XAppA¬pX+˙˙X
14 df-rex pA¬pX+˙˙XppA¬pX+˙˙X
15 13 14 sylibr X+˙˙XAX+˙˙XApA¬pX+˙˙X
16 simplll KHLXA˙˙X=XXpA¬pX+˙˙XKHL
17 simpllr KHLXA˙˙X=XXpA¬pX+˙˙XXA
18 simprl KHLXA˙˙X=XXpA¬pX+˙˙XpA
19 simplrl KHLXA˙˙X=XXpA¬pX+˙˙X˙˙X=X
20 simplrr KHLXA˙˙X=XXpA¬pX+˙˙XX
21 simprr KHLXA˙˙X=XXpA¬pX+˙˙X¬pX+˙˙X
22 eqid K=K
23 eqid joinK=joinK
24 eqid X+˙p=X+˙p
25 22 23 1 2 3 24 pexmidlem6N KHLXApA˙˙X=XX¬pX+˙˙XX+˙p=X
26 22 23 1 2 3 24 pexmidlem7N KHLXApA˙˙X=XX¬pX+˙˙XX+˙pX
27 25 26 jca KHLXApA˙˙X=XX¬pX+˙˙XX+˙p=XX+˙pX
28 16 17 18 19 20 21 27 syl33anc KHLXA˙˙X=XXpA¬pX+˙˙XX+˙p=XX+˙pX
29 nonconne ¬X+˙p=XX+˙pX
30 29 4 2false X+˙p=XX+˙pXX=XXX
31 28 30 sylib KHLXA˙˙X=XXpA¬pX+˙˙XX=XXX
32 31 rexlimdvaa KHLXA˙˙X=XXpA¬pX+˙˙XX=XXX
33 15 32 syl5 KHLXA˙˙X=XXX+˙˙XAX+˙˙XAX=XXX
34 10 33 mpand KHLXA˙˙X=XXX+˙˙XAX=XXX
35 34 necon1bd KHLXA˙˙X=XX¬X=XXXX+˙˙X=A
36 4 35 mpi KHLXA˙˙X=XXX+˙˙X=A