reserve a,b,c,d for Real;
reserve r,s for Real;

theorem
  0 <= d & d <= 1 & (1-d)*a+d*b <= a & (1-d)*a+d*b < b implies d = 0
proof
  assume that
A1: d >= 0 and
A2: d <= 1 and
A3: a >= (1-d)*a + d*b and
A4: b > (1-d)*a + d*b;
  set s = (1-d)*a + d*b;
  assume d <> 0;
  then
A5: d*b > d*s by A1,A4,Lm13;
  1-d >= 0 by A2,Th48;
  then
A6: (1-d)*a >= (1-d)*s by A3,Lm12;
  1*s = (1-d)*s + d*s;
  hence contradiction by A5,A6,Lm8;
end;
