theorem Th26:
  |{p3,p5,p7}| = 0 implies
  |{p1,p3,p5}| * |{p5,p8,p7}| = - |{p1,p5,p7}| * |{p3,p5,p8}|
  proof
    assume
A1: |{p3,p5,p7}| = 0;
A2: |{p3,p5,p7}| = - |{p5,p3,p7}| & |{p5,p8,p3}| = |{p3,p5,p8}| &
      |{p5,p7,p1}| = |{p1,p5,p7}| & |{p1,p3,p5}| = |{p5,p1,p3}|
      by ANPROJ_8:30,Th01; then
A3: |{p5,p3,p7}| = 0 & - |{p1,p3,p5}| = |{p5,p3,p1}| by ANPROJ_8:29,A1;
    |{p5,p8,p3}| * |{p5,p7,p1}| - |{p5,p8,p7}| * |{p5,p3,p1}| +
      |{p5,p8,p1}| * |{p5,p3,p7}| = 0 by ANPROJ_8:28;
    then |{p3,p5,p8}| * |{p1,p5,p7}| = |{p5,p8,p7}| * ( - |{p1,p3,p5}|)
      by A2,A3;
    hence thesis;
  end;
