theorem Th12:
  dom p1 = dom p2 implies dom lmlt(p1,p2) = dom p1
  proof
    assume
    A1: dom p1 = dom p2;
    A2: [:rng p1,rng p2:] c= [:INT,the carrier of V1:] by ZFMISC_1:96;
    A3: rng <:p1,p2:> c= [:rng p1,rng p2:] &
    [:INT,the carrier of V1:] = dom (the lmult of V1)
    by FUNCT_2:def 1,FUNCT_3:51;
    thus dom lmlt(p1,p2) = dom((the lmult of V1)*<:p1,p2:> ) by FUNCOP_1:def 3
    .= dom <:p1,p2:> by A2,A3,RELAT_1:27,XBOOLE_1:1
    .= dom p1 /\ dom p2 by FUNCT_3:def 7
    .= dom p1 by A1;
  end;
