Metamath Proof Explorer


Theorem fh1

Description: Foulis-Holland Theorem. If any 2 pairs in a triple of orthomodular lattice elements commute, the triple is distributive. First of two parts. Theorem 5 of Kalmbach p. 25. (Contributed by NM, 14-Jun-2006) (New usage is discouraged.)

Ref Expression
Assertion fh1 ACBCCCA𝐶BA𝐶CABC=ABAC

Proof

Step Hyp Ref Expression
1 chincl ACBCABC
2 chincl ACCCACC
3 chjcl ABCACCABACC
4 1 2 3 syl2an ACBCACCCABACC
5 4 anandis ACBCCCABACC
6 chjcl BCCCBCC
7 chincl ACBCCABCC
8 6 7 sylan2 ACBCCCABCC
9 chsh ABCCABCS
10 8 9 syl ACBCCCABCS
11 5 10 jca ACBCCCABACCABCS
12 11 3impb ACBCCCABACCABCS
13 12 adantr ACBCCCA𝐶BA𝐶CABACCABCS
14 ledi ACBCCCABACABC
15 14 adantr ACBCCCA𝐶BA𝐶CABACABC
16 incom ABC=BCA
17 16 a1i ACBCACCCABC=BCA
18 chdmj1 ABCACCABAC=ABAC
19 1 2 18 syl2an ACBCACCCABAC=ABAC
20 chdmm1 ACBCAB=AB
21 chdmm1 ACCCAC=AC
22 20 21 ineqan12d ACBCACCCABAC=ABAC
23 19 22 eqtrd ACBCACCCABAC=ABAC
24 17 23 ineq12d ACBCACCCABCABAC=BCAABAC
25 24 3impdi ACBCCCABCABAC=BCAABAC
26 25 adantr ACBCCCA𝐶BA𝐶CABCABAC=BCAABAC
27 inass BCAABAC=BCAABAC
28 cmcm2 ACBCA𝐶BA𝐶B
29 choccl BCBC
30 cmbr3 ACBCA𝐶BAAB=AB
31 29 30 sylan2 ACBCA𝐶BAAB=AB
32 28 31 bitrd ACBCA𝐶BAAB=AB
33 32 biimpa ACBCA𝐶BAAB=AB
34 33 3adantl3 ACBCCCA𝐶BAAB=AB
35 34 adantrr ACBCCCA𝐶BA𝐶CAAB=AB
36 cmcm2 ACCCA𝐶CA𝐶C
37 choccl CCCC
38 cmbr3 ACCCA𝐶CAAC=AC
39 37 38 sylan2 ACCCA𝐶CAAC=AC
40 36 39 bitrd ACCCA𝐶CAAC=AC
41 40 biimpa ACCCA𝐶CAAC=AC
42 41 3adantl2 ACBCCCA𝐶CAAC=AC
43 42 adantrl ACBCCCA𝐶BA𝐶CAAC=AC
44 35 43 ineq12d ACBCCCA𝐶BA𝐶CAABAAC=ABAC
45 inindi AABAC=AABAAC
46 inindi ABC=ABAC
47 44 45 46 3eqtr4g ACBCCCA𝐶BA𝐶CAABAC=ABC
48 47 ineq2d ACBCCCA𝐶BA𝐶CBCAABAC=BCABC
49 27 48 eqtrid ACBCCCA𝐶BA𝐶CBCAABAC=BCABC
50 in12 BCABC=ABCBC
51 49 50 eqtrdi ACBCCCA𝐶BA𝐶CBCAABAC=ABCBC
52 chdmj1 BCCCBC=BC
53 52 ineq2d BCCCBCBC=BCBC
54 chocin BCCBCBC=0
55 6 54 syl BCCCBCBC=0
56 53 55 eqtr3d BCCCBCBC=0
57 56 ineq2d BCCCABCBC=A0
58 chm0 ACA0=0
59 57 58 sylan9eqr ACBCCCABCBC=0
60 59 3impb ACBCCCABCBC=0
61 60 adantr ACBCCCA𝐶BA𝐶CABCBC=0
62 51 61 eqtrd ACBCCCA𝐶BA𝐶CBCAABAC=0
63 26 62 eqtrd ACBCCCA𝐶BA𝐶CABCABAC=0
64 pjoml ABACCABCSABACABCABCABAC=0ABAC=ABC
65 13 15 63 64 syl12anc ACBCCCA𝐶BA𝐶CABAC=ABC
66 65 eqcomd ACBCCCA𝐶BA𝐶CABC=ABAC