reserve a,b,c,x,y,z for Real;

theorem Th55:
  -1 <= a & a <= 1 & -1 <= b & b <= 1 implies (-b)*sqrt(1+a^2) <=
  sqrt(1+b^2) & -sqrt(1+b^2) <= b*sqrt(1+a^2)
proof
  assume that
A1: -1<=a and
A2: a<=1 and
A3: -1<=b and
A4: b<=1;
A5: a^2>=0 by XREAL_1:63;
  then
A6: 1+a^2>=1+0 by XREAL_1:7;
  b^2>=0 by XREAL_1:63;
  then
A7: sqrt(1+b^2)>=0 by Def2;
A8: sqrt(1+a^2)>=0 by A5,Def2;
A9: now
    per cases;
    suppose
      b>=0;
      hence (-b)*sqrt(1+a^2)<=sqrt(1+b^2) by A8,A7;
    end;
    suppose
A10:  b<0;
A11:  (-b)^2>=0 by XREAL_1:63;
      (-b)*sqrt(1+a^2)=sqrt((-b)^2*(1+a^2)) by A5,A10,Th54;
      hence (-b)*sqrt(1+a^2)<=sqrt(1+b^2) by A1,A2,A3,A4,A6,A11,Lm6,Th26;
    end;
  end;
  then -((-b)*sqrt(1+a^2)) >= - sqrt(1+b^2) by XREAL_1:24;
  hence thesis by A9;
end;
