theorem
  a > 0 & b > 0 implies (a/b) #R c = a #R c / b #R c
proof
  assume that
A1: a>0 and
A2: b>0;
  thus (a/b) #R c = (a*b") #R c .= a #R c * (b") #R c by A1,A2,Th78
    .= a #R c * (1/b) #R c
    .= a #R c * (1 / b #R c) by A2,Th79
    .= a #R c * 1 / b #R c
    .= a #R c / b #R c;
end;
