 reserve K,F,E for Field,
         R,S for Ring;

theorem Th15:
   for S being RingExtension of R,
   p being Element of the carrier of (Polynom-Ring R),
   q being Element of the carrier of Polynom-Ring S,
   a being Element of S st p = q holds Ext_eval(p,a) = eval(q,a)
   proof
     let S be RingExtension of R;
     let p be Element of the carrier of (Polynom-Ring R),
     q be Element of the carrier of Polynom-Ring S;
     let a be Element of S;
     assume
A1:  p = q;
A2:  R is Subring of S by Def1;
     per cases;
       suppose q is zero; then
A3:      q = 0_.(S) by UPROOTS:def 5;
         p = 0.(Polynom-Ring S) by A1,A3,POLYNOM3:def 10
         .= 0.(Polynom-Ring R) by Th7
         .= 0_.(R) by POLYNOM3:def 10;
         hence
         Ext_eval(p,a) = 0.S by ALGNUM_1:13 .= eval(q,a) by A3,POLYNOM4:17;
       end;
      suppose
A4:     q is non zero; then
        len q > 0 by UPROOTS:17; then
A5:     len q -' 1 = len q - 1 by XREAL_0:def 2; then
        reconsider lenq = len q - 1 as Element of NAT;
        consider Fp being FinSequence of S such that
A6:     Ext_eval(p,a) = Sum Fp & len Fp = len p &
        for n being Element of NAT st n in dom Fp holds
        Fp.n = In(p.(n-'1),S) * (power S).(a,n-'1) by ALGNUM_1:def 1;
        consider Fq being FinSequence of the carrier of S such that
A7:     eval(q,a) = Sum Fq & len Fq = len q &
        for n being Element of NAT st n in dom Fq holds
        Fq.n = q.(n-'1) * (power S).(a,n-'1) by POLYNOM4:def 2;
A8:     now let i be Nat;
          assume i >= len q; then
          q.i = 0.S by ALGSEQ_1:8;
          hence p.i = 0.R by A1,A2,C0SP1:def 3;
        end;
        now let m be Nat;
          assume
A9:       m is_at_least_length_of p;
          now assume len q > m; then
          lenq + 1 > m; then
          lenq >= m by NAT_1:13; then
          p.(len q-'1) = 0.R by A5,A9,ALGSEQ_1:def 2; then
A10:       q.(len q-'1) = 0.S by A1,A2,C0SP1:def 3;
          0 + 1 < len q + 1 by A4,XREAL_1:8,UPROOTS:17; then
          len q >= 1 by NAT_1:13; then
          (len q-'1) + 1 = len q by XREAL_1:235;
          hence contradiction by A10,ALGSEQ_1:10;
        end;
        hence len q <= m;
      end; then
      len p = len q by A8,ALGSEQ_1:def 3,ALGSEQ_1:def 2; then
A11:  dom Fq = Seg(len p) by A7,FINSEQ_1:def 3
      .= dom Fp by A6,FINSEQ_1:def 3;
      for n being Nat st n in dom Fp holds Fq.n = Fp.n
      proof
        let n be Nat;
        assume
A12:    n in dom Fp;
        hence Fp.n = In(p.(n-'1),S) * (power S).(a,n-'1) by A6
        .= q.(n-'1) * (power S).(a,n-'1) by A1
        .= Fq.n by A7,A11,A12;
      end;
      hence thesis by A6,A7,A11,FINSEQ_1:13;
    end;
  end;
