 reserve n,m for Nat,
      o for object,
      p for pair object,
      x,y,z for Surreal;

theorem
  0_No < x implies (sqrt x)" == sqrt (x")
proof
  assume
A1: 0_No < x;
  then reconsider a = x as positive Surreal by SURREALI:def 8;
  set S=sqrt a,T = sqrt (a");
A2:not S ==0_No & not T == 0_No by SURREALI:def 8;
A3:0_No <= a" by SURREALI:def 8;
  0_No <= x by A1;
  then S*S*(T*T) == a*(T*T) == a*a" by A3,Th21,SURREALR:54;
  then
A4: (S*S)*(T*T) == a*a" by SURREALO:4;
  not a == 0_No by A1;
  then a*a" == 1_No by SURREALI:33;
  then
A5: (S*S)*(T*T) == 1_No by A4,SURREALO:4;
A6: 0_No <= 1_No by SURREALI:def 8;
A7:-1_No <= -0_No by SURREALR:10,SURREALI:def 8;
  (S*S)*(T*T) == (S*S)*T*T == S*(S*T)*T by SURREALR:69,54;
  then (S*S)*(T*T) == S*(S*T)*T == (S*T)*(S*T)
  by SURREALO:4,SURREALR:69;
  then (S*S)*(T*T) == (S*T)*(S*T) by SURREALO:4;
  then (S*T)*(S*T) == 1_No by A5,SURREALO:4;
  then S*T == sqrt 1_No or S*T == - sqrt 1_No by A6,Th28;
  hence thesis by A2,A7,SURREALI:def 8,42,SURREALO:4;
end;
