theorem Th115:
  A///B = {c1/c2: c1 in A & c2 in B}
proof
  thus A///B c= {c1/c2: c1 in A & c2 in B}
  proof
    let e be object;
    assume e in A///B;
    then consider c1,c2 such that
A1: e = c1*c2 and
A2: c1 in A and
A3: c2 in B"";
    e = c1/(c2") & c2" in B by A1,A3,Th29;
    hence thesis by A2;
  end;
  let e be object;
  assume e in {c1/c2: c1 in A & c2 in B};
  then consider c1,c2 such that
A4: e = c1/c2 & c1 in A and
A5: c2 in B;
  c2" in B"" by A5;
  hence thesis by A4;
end;
