theorem
  (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 = n2 & m1 = m2
proof
A1: 2|^n1 <> 0 by Th87;
  assume
A2: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  then
A3: n2 <= n1 by Lm5;
  n1 <= n2 by A2,Lm5;
  hence n1 = n2 by A3,XXREAL_0:1;
  then 2*m1+1 = 2*m2+1 by A2,A1,XCMPLX_1:5;
  hence thesis;
end;
