theorem Th20:
  (i in dom p implies [#](p,d).i = p.i) & (not i in dom p implies
  [#](p,d).i = d)
proof
  thus i in dom p implies [#](p,d).i = p.i by FUNCT_4:13;
A1: i in NAT by ORDINAL1:def 12;
  assume not i in dom p;
  hence [#](p,d).i = (NAT --> d).i by FUNCT_4:11
    .= d by A1,FUNCOP_1:7;
end;
