theorem Th100:
  A\iffB\orC in F & B\iffD in F implies A\iffD\orC in F
  proof
    assume A1: A\iffB\orC in F;
    then A2: B\orC\iffA in F by Th90;
    assume B\iffD in F;
    then B\impD in F & D\impB in F & C\impC in F by Th43,Th34;
    then B\orC\impD\orC in F & D\orC\impB\orC in F by Th59;
    then A\impD\orC in F & D\orC\impA in F by A1,A2,Th92,Th93;
    hence A\iffD\orC in F by Th43;
  end;
