
theorem Th48: ::: FIB_NUM2:48 revised
  for k,n being Nat st k <> 1 & n <> 1 holds
    Fib k = Fib n implies k = n
proof
    let k, n be Nat;
    assume that
A0: k <> 1 & n <> 1 and
A1: Fib k = Fib n and
A2: k <> n;
   per cases by A2,XXREAL_0:1;
    suppose A3: k < n;
      per cases by A0,XXREAL_0:1;
      suppose k < 1;
        then k = 0 by NAT_1:14;
        then Fib k = 0 by PRE_FF:1;
        hence contradiction by A1,A3,LemmaFib0;
      end;
      suppose k > 1;
        hence contradiction by A1,A3,Th46;
      end;
    end;
    suppose A4: n < k;
      per cases by A0,XXREAL_0:1;
      suppose n < 1;
        then n = 0 by NAT_1:14;
        then Fib n = 0 by PRE_FF:1;
        hence contradiction by A1,A4,LemmaFib0;
      end;
      suppose n > 1;
        hence contradiction by A1,A4,FIB_NUM2:46;
      end;
    end;
  end;
