theorem Th98:
  A\iffB\andC in F & B\iffD in F implies A\iffD\andC in F
  proof
    assume A1: A\iffB\andC in F;
    then A2: B\andC\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\andC\impD\andC in F & D\andC\impB\andC in F by Th72;
    then A\impD\andC in F & D\andC\impA in F by A1,A2,Th92,Th93;
    hence A\iffD\andC in F by Th43;
  end;
