
theorem TT2:
for L being non empty 1-sorted
for f being bijective Function of L,L
for n,m being Nat holds f`^(n+1) = f`^(m+1) iff f`^n = f`^m
proof
let L be non empty 1-sorted, f be bijective Function of L,L;
let n,m be Nat;
H: f is onto;
A: now assume f`^(n+1) = f`^(m+1); then
   ((f`^n) * f) * f" = f`^(m+1) * f" by T3
      .= ((f`^m) * f) * f" by T3
      .= (f`^m) * (f * f") by T2
      .= (f`^m) * (id L) by H,FUNCT_1:39
      .= f`^m;
   hence f`^m = (f`^n) * (f * f") by T2
      .= (f`^n) * (id L) by H,FUNCT_1:39
      .= f`^n;
   end;
now assume f`^n = f`^m;
  hence f`^(n+1) = ((f`^m) * f) by T3 .= f`^(m+1) by T3;
  end;
hence thesis by A;
end;
