theorem
  L1.v <> L2.v implies ((r*L1+(1-r)*L2).v = s iff r = (L2.v-s)/(L2.v-L1.v))
 proof
  set u1=L1.v,u2=L2.v;
  A1: (r*L1+(1-r)*L2).v=(r*L1).v+((1-r)*L2).v by RLVECT_2:def 10
   .=r*u1+((1-r)*L2).v by RLVECT_2:def 11
   .=r*u1+(-r+1)*u2 by RLVECT_2:def 11
   .=r*(u1-u2)+u2;
  assume A2: u1<>u2;
  then A3: u1-u2<>0;
  A4: u2-u1<>0 by A2;
  hereby assume(r*L1+(1-r)*L2).v=s;
   then r*(u2-u1)=(u2-s)*1 by A1;
   then r/1=(u2-s)/(u2-u1) by A4,XCMPLX_1:94;
   hence r=(u2-s)/(u2-u1);
  end;
  assume r=(u2-s)/(u2-u1);
  hence (r*L1+(1-r)*L2).v=(u2-s)/(-(u1-u2))*(u1-u2)+u2 by A1
   .=(-(u2-s))/(u1-u2)*(u1-u2)+u2 by XCMPLX_1:192
   .=-(u2-s)+u2 by A3,XCMPLX_1:87
   .=s;
 end;
