Metamath Proof Explorer


Theorem atcvrj0

Description: Two atoms covering the zero subspace are equal. ( atcv1 analog.) (Contributed by NM, 29-Nov-2011)

Ref Expression
Hypotheses atcvrj0.b B = Base K
atcvrj0.j ˙ = join K
atcvrj0.z 0 ˙ = 0. K
atcvrj0.c C = K
atcvrj0.a A = Atoms K
Assertion atcvrj0 K HL X B P A Q A X C P ˙ Q X = 0 ˙ P = Q

Proof

Step Hyp Ref Expression
1 atcvrj0.b B = Base K
2 atcvrj0.j ˙ = join K
3 atcvrj0.z 0 ˙ = 0. K
4 atcvrj0.c C = K
5 atcvrj0.a A = Atoms K
6 breq1 X = 0 ˙ X C P ˙ Q 0 ˙ C P ˙ Q
7 6 biimpd X = 0 ˙ X C P ˙ Q 0 ˙ C P ˙ Q
8 7 adantl K HL X B P A Q A X = 0 ˙ X C P ˙ Q 0 ˙ C P ˙ Q
9 2 3 4 5 atcvr0eq K HL P A Q A 0 ˙ C P ˙ Q P = Q
10 9 3adant3r1 K HL X B P A Q A 0 ˙ C P ˙ Q P = Q
11 10 adantr K HL X B P A Q A X = 0 ˙ 0 ˙ C P ˙ Q P = Q
12 8 11 sylibd K HL X B P A Q A X = 0 ˙ X C P ˙ Q P = Q
13 12 ex K HL X B P A Q A X = 0 ˙ X C P ˙ Q P = Q
14 13 com23 K HL X B P A Q A X C P ˙ Q X = 0 ˙ P = Q
15 14 3impia K HL X B P A Q A X C P ˙ Q X = 0 ˙ P = Q
16 oveq1 P = Q P ˙ Q = Q ˙ Q
17 16 breq2d P = Q X C P ˙ Q X C Q ˙ Q
18 17 biimpac X C P ˙ Q P = Q X C Q ˙ Q
19 simpr3 K HL X B P A Q A Q A
20 2 5 hlatjidm K HL Q A Q ˙ Q = Q
21 19 20 syldan K HL X B P A Q A Q ˙ Q = Q
22 21 breq2d K HL X B P A Q A X C Q ˙ Q X C Q
23 hlatl K HL K AtLat
24 23 adantr K HL X B P A Q A K AtLat
25 simpr1 K HL X B P A Q A X B
26 eqid K = K
27 1 26 3 4 5 atcvreq0 K AtLat X B Q A X C Q X = 0 ˙
28 24 25 19 27 syl3anc K HL X B P A Q A X C Q X = 0 ˙
29 28 biimpd K HL X B P A Q A X C Q X = 0 ˙
30 22 29 sylbid K HL X B P A Q A X C Q ˙ Q X = 0 ˙
31 18 30 syl5 K HL X B P A Q A X C P ˙ Q P = Q X = 0 ˙
32 31 expd K HL X B P A Q A X C P ˙ Q P = Q X = 0 ˙
33 32 3impia K HL X B P A Q A X C P ˙ Q P = Q X = 0 ˙
34 15 33 impbid K HL X B P A Q A X C P ˙ Q X = 0 ˙ P = Q