
theorem
  Fib(0) = 0 & Fib(1) = 1 &
  for n being Nat holds Fib((n+1)+1) = Fib(n) + Fib(n+1)
proof
A1: Fib.0 = [0,1] by Def1;
A3: for n being Nat holds Fib.(n+1) = [(Fib.n)`2, (Fib.n)`1 + (Fib.n)`2]
    by Def1;
  thus Fib (0) = [0, 1]`1 by Def1
    .= 0;
  thus Fib (1) = (Fib.(0+1))`1
    .= ([ (Fib.0)`2, (Fib.0)`1 + (Fib.0)`2 ])`1 by A3
    .= 1 by A1;
  let n be Nat;
A4: (Fib.(n+1))`1 = [ (Fib.n)`2, (Fib.n)`1 + (Fib.n)`2 ]`1 by A3
    .= (Fib.n)`2;
  thus Fib((n+1)+1) = [ (Fib.(n+1))`2, (Fib.(n+1))`1 + (Fib.(n+1))`2 ]`1 by A3
    .= [ (Fib.n)`2, (Fib.n)`1 + (Fib.n)`2 ]`2 by A3
    .= Fib(n) + Fib(n+1) by A4;
end;
