Metamath Proof Explorer


Theorem dpjdisj

Description: The two subgroups that appear in dpjval are disjoint. (Contributed by Mario Carneiro, 26-Apr-2016)

Ref Expression
Hypotheses dpjfval.1 φGdomDProdS
dpjfval.2 φdomS=I
dpjlem.3 φXI
dpjdisj.0 0˙=0G
Assertion dpjdisj φSXGDProdSIX=0˙

Proof

Step Hyp Ref Expression
1 dpjfval.1 φGdomDProdS
2 dpjfval.2 φdomS=I
3 dpjlem.3 φXI
4 dpjdisj.0 0˙=0G
5 1 2 3 dpjlem φGDProdSX=SX
6 5 ineq1d φGDProdSXGDProdSIX=SXGDProdSIX
7 1 2 dprdf2 φS:ISubGrpG
8 disjdif XIX=
9 8 a1i φXIX=
10 undif2 XIX=XI
11 3 snssd φXI
12 ssequn1 XIXI=I
13 11 12 sylib φXI=I
14 10 13 eqtr2id φI=XIX
15 eqid CntzG=CntzG
16 7 9 14 15 4 dmdprdsplit φGdomDProdSGdomDProdSXGdomDProdSIXGDProdSXCntzGGDProdSIXGDProdSXGDProdSIX=0˙
17 1 16 mpbid φGdomDProdSXGdomDProdSIXGDProdSXCntzGGDProdSIXGDProdSXGDProdSIX=0˙
18 17 simp3d φGDProdSXGDProdSIX=0˙
19 6 18 eqtr3d φSXGDProdSIX=0˙