theorem Th4:
  (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1) implies n1 = n2 & m1 = m2
proof
A1: 2|^n1 <> 0 by Th3;
  assume
A2: (2|^n1)*(2*m1+1) = (2|^n2)*(2*m2+1);
  then n1 <= n2 & n2 <= n1 by Lm1;
  hence n1 = n2 by XXREAL_0:1;
  then 2*m1+1 = 2*m2+1 by A2,A1,XCMPLX_1:5;
  hence thesis;
end;
