theorem Th84:
  a>0 & a<=1 & c>=b implies a #R c <= a #R b
proof
  assume that
A1: a>0 and
A2: a<=1 and
A3: c>=b;
  1/a>=1 by A1,A2,Lm4,XREAL_1:85;
  then (1/a) #R c >= (1/a) #R b by A3,Th82;
  then 1/a #R c >= (1/a) #R b by A1,Th79;
  then
A4: 1/a #R c >= 1/a #R b by A1,Th79;
  a #R b > 0 by A1,Th81;
  hence thesis by A4,XREAL_1:89;
end;
