
theorem Th5:
for F be FinSequence of COMPLEX, x be Element of COMPLEX holds
    Re (F^<* x *>) = (Re F)^<* Re x *>
proof
let F be FinSequence of COMPLEX,
    x be Element of COMPLEX;
set F1=Re (F^<* x *>);
set F2=(Re F)^<* Re x *>;
A1: dom F = dom (Re F) by COMSEQ_3:def 3;
A2: Seg len F = dom F by FINSEQ_1:def 3;
A3: len <* Re x *> = 1 by FINSEQ_1:39;
A4: dom F1 = dom (F^<* x *>) by COMSEQ_3:def 3
 .= Seg (len F + len <* x *>) by FINSEQ_1:def 7
 .= Seg (len (Re F) + len <* x *>) by A1,A2,FINSEQ_1:def 3
 .= Seg (len (Re F) + len <* Re x *>) by A3,FINSEQ_1:39
 .= dom F2 by FINSEQ_1:def 7;
  now let k be Nat;
    assume A5: k in dom F1; then
    k in dom (F^<*x*>) by COMSEQ_3:def 3; then
  A6:k in Seg len((F^<*x*>)) by FINSEQ_1:def 3; then
    k in Seg (len F + len <*x*>) by FINSEQ_1:22; then
  A7:1 <=k & k <= len F + len <*x*> by FINSEQ_1:1;
    now per cases;
      suppose A8: k = len F + 1;
      thus F1.k = Re( (F^<* x *>).k ) by A5,COMSEQ_3:def 3
               .= Re x by A8,FINSEQ_1:42
               .= F2.(len (Re F) + 1) by FINSEQ_1:42
               .= F2.k by A8,A1,A2,FINSEQ_1:def 3;
      end;
      suppose A9: k <> len F + 1;
      k <= len F + 1 by A7,FINSEQ_1:39; then
      k < len F + 1 by A9,XXREAL_0:1; then
      1 <= k & k <= len F by A6,FINSEQ_1:1,NAT_1:13; then
      k in Seg (len F); then
  A10:k in dom F by FINSEQ_1:def 3; then
  A11:k in dom (Re F) by COMSEQ_3:def 3;
      thus F1.k = Re( (F^<* x *>).k ) by A5,COMSEQ_3:def 3
               .= Re (F.k) by A10,FINSEQ_1:def 7
               .=(Re F).k by A11,COMSEQ_3:def 3
               .=F2.k by A11,FINSEQ_1:def 7;
      end;
    end;
    hence F1.k = F2.k;
  end;
hence Re (F^<* x *>) = (Re F)^<* Re x *> by A4,FINSEQ_1:13;
end;
