Metamath Proof Explorer


Theorem llnexatN

Description: Given an atom on a line, there is another atom whose join equals the line. (Contributed by NM, 26-Jun-2012) (New usage is discouraged.)

Ref Expression
Hypotheses llnexat.l ˙=K
llnexat.j ˙=joinK
llnexat.a A=AtomsK
llnexat.n N=LLinesK
Assertion llnexatN KHLXNPAP˙XqAPqX=P˙q

Proof

Step Hyp Ref Expression
1 llnexat.l ˙=K
2 llnexat.j ˙=joinK
3 llnexat.a A=AtomsK
4 llnexat.n N=LLinesK
5 simp1 KHLXNPAKHL
6 simp3 KHLXNPAPA
7 simp2 KHLXNPAXN
8 5 6 7 3jca KHLXNPAKHLPAXN
9 eqid K=K
10 1 9 3 4 atcvrlln2 KHLPAXNP˙XPKX
11 8 10 sylan KHLXNPAP˙XPKX
12 simpl1 KHLXNPAP˙XKHL
13 simpl3 KHLXNPAP˙XPA
14 eqid BaseK=BaseK
15 14 3 atbase PAPBaseK
16 13 15 syl KHLXNPAP˙XPBaseK
17 simpl2 KHLXNPAP˙XXN
18 14 4 llnbase XNXBaseK
19 17 18 syl KHLXNPAP˙XXBaseK
20 14 1 2 9 3 cvrval3 KHLPBaseKXBaseKPKXqA¬q˙PP˙q=X
21 12 16 19 20 syl3anc KHLXNPAP˙XPKXqA¬q˙PP˙q=X
22 simpll1 KHLXNPAP˙XqAKHL
23 hlatl KHLKAtLat
24 22 23 syl KHLXNPAP˙XqAKAtLat
25 simpr KHLXNPAP˙XqAqA
26 simpll3 KHLXNPAP˙XqAPA
27 1 3 atncmp KAtLatqAPA¬q˙PqP
28 24 25 26 27 syl3anc KHLXNPAP˙XqA¬q˙PqP
29 28 anbi1d KHLXNPAP˙XqA¬q˙PP˙q=XqPP˙q=X
30 necom qPPq
31 eqcom P˙q=XX=P˙q
32 30 31 anbi12i qPP˙q=XPqX=P˙q
33 29 32 bitrdi KHLXNPAP˙XqA¬q˙PP˙q=XPqX=P˙q
34 33 rexbidva KHLXNPAP˙XqA¬q˙PP˙q=XqAPqX=P˙q
35 21 34 bitrd KHLXNPAP˙XPKXqAPqX=P˙q
36 11 35 mpbid KHLXNPAP˙XqAPqX=P˙q