
theorem TT3:
for L being non empty 1-sorted
for f being bijective Function of L,L
for n,m,k being Nat st f`^n = f`^m & k = n-m holds f`^k = f`^0
proof
let L be non empty 1-sorted, f be bijective Function of L,L;
let n,m,k be Nat;
assume AS: f`^n = f`^m & k = n-m;
defpred P[Nat] means for n,k being Nat holds
  f`^n = f`^($1) & k = n-($1) implies f`^k = f`^0;
IA: P[0];
IS: now let m be Nat;
    assume IV: P[m];
    now let n,k be Nat;
      assume AS: f`^n = f`^(m+1) & k = n-(m+1); then
      n - (m + 1) + m >= 0 + m by XREAL_1:6; then
      reconsider n1 = n - 1 as Element of NAT by INT_1:3;
      reconsider k1 = n1 - m as Nat by AS;
      n = n1 + 1;
      then f`^n1 = f`^m by AS,TT2;
      hence f`^0 = f`^k1 by IV .= f`^k by AS;
      end;
    hence P[m+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis by AS;
end;
