reserve A,X,X1,X2,Y,Y1,Y2 for set, a,b,c,d,x,y,z for object;
reserve P,P1,P2,Q,R,S for Relation;

theorem
  rng R misses dom P implies R*P = {}
proof
  assume
A1: rng R /\ dom P = {};
  assume R*P <> {};
  then consider x,z such that
A2: [x,z] in R*P;
  consider y such that
A3: [x,y] in R & [y,z] in P by A2,Def6;
  y in rng R & y in dom P by A3,XTUPLE_0:def 12,def 13;
  hence contradiction by A1,XBOOLE_0:def 4;
end;
