theorem
  A c= B & C c= D implies A///C c= B///D
proof
  assume
A1: A c= B & C c= D;
  let z;
A2: A///C = {c1/c2: c1 in A & c2 in C} by Th115;
  assume z in A///C;
  then ex c,c1 st z = c/c1 & c in A & c1 in C by A2;
  hence thesis by A1,Th116;
end;
