theorem Th22:
  |{p1,p6,p8}| = 0 implies
  |{p1,p2,p6}| * |{p3,p6,p8}| = |{p1,p3,p6}| * |{p2,p6,p8}|
  proof
    assume
A1: |{p1,p6,p8}| = 0;
A2: |{p1,p6,p8}| = - |{p6,p1,p8}| & |{p6,p8,p3}| = |{p3,p6,p8}| &
    |{p6,p1,p3}| = |{p1,p3,p6}| & |{p6,p8,p2}| = |{p2,p6,p8}| &
    |{p1,p2,p6}| = |{p6,p1,p2}| by ANPROJ_8:30,Th01; then
A3: |{p6,p1,p8}| = 0 by A1;
    |{p6,p1,p8}| * |{p6,p2,p3}| - |{p6,p1,p2}| * |{p6,p8,p3}| +
      |{p6,p1,p3}| * |{p6,p8,p2}| = 0 by ANPROJ_8:28;
    hence thesis by A2,A3;
  end;
