 reserve R for domRing;
 reserve p for odd prime Nat, m for positive Nat;
 reserve g for non zero Polynomial of INT.Ring;

theorem Th10:
  for k be non zero Nat,p be odd prime Nat holds
  (x.(k,p))^<* (tau(k+1))|^p *> = x.(k+1,p) &
  Product(x.(k+1,p)) = Product((x.(k,p)))*((tau(k+1))|^p)
    proof
      let k be non zero Nat, p be odd prime Nat;
      (x.(k,p))^<* (tau(k+1))|^p *> = x.(k+1,p)
      proof
A2:     dom (x.(k,p)) = Seg len(x.(k,p)) by FINSEQ_1:def 3
        .= Seg k by Def2;
A3:     len x.(k+1,p) = k+1 by Def2; then
A5:     dom x.(k+1,p) = Seg (k+1) by FINSEQ_1:def 3;
        len((x.(k,p))^<* (tau(k+1))|^p *>)
        = len (x.(k,p)) + 1 by FINSEQ_2:16
        .= k +1 by Def2; then
A6:     dom ((x.(k,p))^<* (tau(k+1))|^p *>) = Seg (k+1) by FINSEQ_1:def 3;
        for i be Nat st i in dom x.(k+1,p) holds
        (x.(k+1,p)).i = ((x.(k,p))^<* (tau(k+1))|^p *>).i
        proof
          let i be Nat;
          assume i in dom x.(k+1,p); then
          i in Seg (k+1) by A3,FINSEQ_1:def 3; then
          i in Seg k \/ { k+1 } by FINSEQ_1:9; then
          per cases by XBOOLE_0:def 3;
            suppose
A9:           i in Seg k;
              k < k+1 by NAT_1:19; then
              Seg k c= Seg(k+1) by FINSEQ_1:5; then
              i in Seg (k+1) by A9; then
A10:          i in dom (x.(k+1,p)) by A3,FINSEQ_1:def 3;
              ((x.(k,p))^<* (tau(k+1))|^p *>).i = (x.(k,p)).i
                by A9,A2,FINSEQ_1:def 7
              .= (tau(i))|^p by A9,A2,Def2 .= (x.(k+1,p)).i by A10,Def2;
              hence thesis;
            end;
            suppose
              i in {k+1}; then
A12:          i = k+1 by TARSKI:def 1;
              len (x.(k,p)) = k by Def2;
              hence thesis by A12,A5,FINSEQ_1:3,Def2;
            end;
          end;
          hence thesis by A5,A6;
        end;
       hence thesis by GROUP_4:6;
     end;
