theorem
  dom ((x --> y)+*(x .-->z)+*(succ x .-->z)) = succ succ x
proof
  thus dom ((x --> y)+*(x .-->z)+*(succ x .-->z))
      = dom ((x --> y)+*(x .-->z)) \/ dom(succ x .-->z) by Def1
     .= dom (x --> y) \/ dom (x .-->z) \/ dom(succ x .-->z) by Def1

    .= x \/ dom (x .-->z) \/ dom(succ x .-->z)
    .= x \/ {x} \/ dom(succ x .-->z)
    .= x \/ {x} \/ {succ x}
    .= succ x \/ {succ x} by ORDINAL1:def 1
    .= succ succ x by ORDINAL1:def 1;
end;
