Metamath Proof Explorer


Theorem dihsumssj

Description: The subspace sum of two isomorphisms of lattice elements is less than the isomorphism of their lattice join. (Contributed by NM, 23-Sep-2014)

Ref Expression
Hypotheses dihsumssj.b B = Base K
dihsumssj.h H = LHyp K
dihsumssj.j ˙ = join K
dihsumssj.u U = DVecH K W
dihsumssj.p ˙ = LSSum U
dihsumssj.i I = DIsoH K W
dihsumssj.k φ K HL W H
dihsumssj.x φ X B
dihsumssj.y φ Y B
Assertion dihsumssj φ I X ˙ I Y I X ˙ Y

Proof

Step Hyp Ref Expression
1 dihsumssj.b B = Base K
2 dihsumssj.h H = LHyp K
3 dihsumssj.j ˙ = join K
4 dihsumssj.u U = DVecH K W
5 dihsumssj.p ˙ = LSSum U
6 dihsumssj.i I = DIsoH K W
7 dihsumssj.k φ K HL W H
8 dihsumssj.x φ X B
9 dihsumssj.y φ Y B
10 eqid Base U = Base U
11 eqid joinH K W = joinH K W
12 1 2 6 4 10 dihss K HL W H X B I X Base U
13 7 8 12 syl2anc φ I X Base U
14 1 2 6 4 10 dihss K HL W H Y B I Y Base U
15 7 9 14 syl2anc φ I Y Base U
16 2 4 10 5 11 7 13 15 djhsumss φ I X ˙ I Y I X joinH K W I Y
17 1 3 2 6 11 djhlj K HL W H X B Y B I X ˙ Y = I X joinH K W I Y
18 7 8 9 17 syl12anc φ I X ˙ Y = I X joinH K W I Y
19 16 18 sseqtrrd φ I X ˙ I Y I X ˙ Y