
theorem poly1:
for R being Ring,
    p being Polynomial of R
for a being Element of R holds a * p = (a|R) *' p
proof
let R be Ring, p be Polynomial of R; let a be Element of R;
set q = (a|R) *' p;
A: now let x be object;
   assume x in dom(a * p);
   then reconsider i = x as Element of NAT;
   consider F being FinSequence of the carrier of R such that
   A1: len F = i+1 & q.i = Sum F &
       for k being Element of NAT st k in dom F
       holds F.k = (a|R).(k-'1) * p.(i+1-'k) by POLYNOM3:def 9;
   A2: now let k be Element of NAT;
       assume A3: k in dom F & k <> 1; then
       k in Seg(i+1) by A1,FINSEQ_1:def 3; then
       1 <= k <= i + 1 by FINSEQ_1:1; then
       A6: k -' 1 = k - 1 by XREAL_1:233;
       A4: k - 1 <> 0 by A3;
       thus F/.k = F.k by A3,PARTFUN1:def 6
                .= (a|R).(k-'1) * p.(i+1-'k) by A1,A3
                .= 0.R * p.(i+1-'k) by A6,A4,poly0
                .= 0.R;
       end;
   I: i+1-'1 = i+1-1 by NAT_1:11,XREAL_1:233;
   1 <= i + 1 by NAT_1:11; then
   1 in Seg(i+1); then
   A3: 1 in dom F by A1,FINSEQ_1:def 3; then
   A4: F.1 = (a|R).(1-'1) * p.(i+1-'1) by A1
          .= (a|R).0 * p.(i+1-'1) by NAT_2:8
          .= a * p.i by I,poly0;
   Sum F = F/.1 by A2,A3,POLYNOM2:3 .= F.1 by A3,PARTFUN1:def 6;
   hence (a * p).x = q.x by A4,A1,POLYNOM5:def 4;
   end;
dom(a * p) = NAT by FUNCT_2:def 1 .= dom q by FUNCT_2:def 1;
hence thesis by A;
end;
