reserve a,b,n for Element of NAT;

theorem Th11:
  Lucas(0) = 2 & Lucas(1) = 1 &
  for n being Nat holds Lucas((n+1)+1) = Lucas(n) + Lucas(n+1)
proof
  set L = Lucas;
A1: L.0 = [2,1] by Def1;
  thus Lucas (0) = [2, 1]`1 by Def1
    .= 2;
  thus Lucas (1) = (L.(0+1))`1
    .= ([ (L.0)`2, (L.0)`1 + (L.0)`2 ])`1 by Def1
    .= 1 by A1;
  let n be Nat;
A3: (L.(n+1))`1 = [ (L.n)`2, (L.n)`1 + (L.n)`2 ]`1 by Def1
    .= (L.n)`2;
  thus Lucas((n+1)+1) = [ (L.(n+1))`2, (L.(n+1))`1 + (L.(n+1))`2 ]`1 by Def1
    .= [ (L.n)`2, (L.n)`1 + (L.n)`2 ]`2 by Def1
    .= Lucas(n) + Lucas(n+1) by A3;
end;
