theorem Th12:
  dom p1 = dom p2 implies dom lmlt(p1,p2) = dom p1
proof
  assume
A1: dom p1 = dom p2;
  rng p1 c= the carrier of K & rng p2 c= the carrier of V1 by FINSEQ_1:def 4;
  then
A2: [:rng p1,rng p2:] c= [:the carrier of K,the carrier of V1:] by ZFMISC_1:96;
  rng <:p1,p2:> c= [:rng p1,rng p2:] & [:the carrier of K,the carrier of
  V1:] = dom (the lmult of V1) by FUNCT_2:def 1,FUNCT_3:51;
  hence dom lmlt(p1,p2) = dom <:p1,p2:> by A2,RELAT_1:27,XBOOLE_1:1
    .= dom p1 /\ dom p2 by FUNCT_3:def 7
    .= dom p1 by A1;
end;
