theorem Th34:
  seq"(#)seq1"=(seq(#)seq1)"
proof
  now
    let n be Element of NAT;
    thus ((seq")(#)(seq1")).n=((seq").n)*((seq1").n) by Th8
      .=((seq").n)*(seq1.n)" by VALUED_1:10
      .=(seq.n)"*(seq1.n)" by VALUED_1:10
      .=((seq.n)*(seq1.n))" by XCMPLX_1:204
      .=((seq(#)seq1).n)" by Th8
      .=((seq(#)seq1)").n by VALUED_1:10;
  end;
  hence thesis by FUNCT_2:63;
end;
