theorem Th36:
  x in rng p & k in dom(p -| x) implies p.k = (p -| x).k
proof
  assume that
A1: x in rng p and
A2: k in dom(p -| x);
  ex n st n = x..p - 1 & p | Seg n = p -| x by A1,Def5;
  hence thesis by A2,FUNCT_1:47;
end;
