theorem Th108:
  X <> {} & rng f c= dom p implies (p/*f).x = p/.(f.x)
proof
  assume that
A1: X <> {} and
A2: rng f c= dom p;
A3: f.x in rng f by A1,Th4;
  thus (p/*f).x = p.(f.x) by A1,A2,Th107
    .= p/.(f.x) by A2,A3,PARTFUN1:def 6;
end;
