
theorem Th1:
  for x,y be Nat st x+y = 1 holds (x = 1 & y = 0) or (x = 0 & y = 1)
  proof
    let x, y be Nat;
    assume A1: x+y = 1;
    x <= 1
    proof
      assume not x <= 1; then
      1 + 0 < x + y by XREAL_1:8;
      hence contradiction by A1;
    end;
    then x=0 or x = 1 by NAT_1:25;
    hence (x = 1 & y = 0) or (x = 0 & y = 1) by A1;
  end;
