theorem Th148:
  A\iffB in H implies \Cap(M,A)\iff\Cap(M,B) in H
  proof
    assume A\iffB in H;
    then A\impB in H & B\impA in H by Th43;
    then \Cap(M,A)\imp\Cap(M,B) in H & \Cap(M,B)\imp\Cap(M,A) in H by Def43;
    hence \Cap(M,A)\iff\Cap(M,B) in H by Th43;
  end;
