
theorem fresh:
for p being Prime
for R being p-characteristic commutative Ring
for a,b being Element of R holds (a + b)|^p = a|^p + b|^p
proof
let p be Prime, R be p-characteristic commutative Ring,
    a,b be Element of R;
set F = (a,b) In_Power p;
len F = p + 1 by BINOM:def 7; then
A0: F.1 = a|^p & F.(len F) = b|^p by BINOM:23,24;
A1: now let i be Nat;
    assume B0: 1 < i & i < len F; then
    i in Seg(len F) by FINSEQ_1:1; then
    B1: i in dom F by FINSEQ_1:def 3;
    reconsider m = i - 1 as Nat by B0;
    i - 1 < len F - 1 by B0,XREAL_1:9; then
    B2: m < (p + 1) - 1 by BINOM:def 7; then
    p - m in NAT by INT_1:5; then
    reconsider l = p - m as Nat;
    1 - 1 < m by B0,XREAL_1:9; then
    B3: 0 + 1 <= m by INT_1:7;
    thus F.i = F/.i by B1,PARTFUN1:def 6
            .= (p choose m) * a|^l * b|^m by B1,BINOM:def 7
            .= 0.R * b|^m by B3,B2,Lm2,NEWTON02:119
            .= 0.R;
   end;
consider fp being sequence of the carrier of R such that
A2: Sum F = fp.(len F) and
A3: fp.0 = 0.R and
A4: for j being Nat, v being Element of R
    st j < len F & v = F.(j + 1)
    holds fp.(j + 1) = fp.j + v by RLVECT_1:def 12;
defpred P[Element of NAT] means
  ($1 = 0 & fp.($1) = 0.R) or
  (0 < $1 & $1 < len F & fp.($1) = a|^p) or
  ($1 = len F & fp.($1) = a|^p + b|^p);
IA: P[0] by A3;
IS: now let j be Element of NAT;
    assume C1: 0 <= j & j < len F;
    assume C2: P[j];
    per cases by XXREAL_0:1;
    suppose D1: j = 0 & j < len F - 1; then
      j < len F; then
      D2: fp.(j+1) = fp.0 + a|^p by D1,A0,A4;
      j + 1 < (len F - 1) + 1 by D1,XREAL_1:6;
      hence P[j+1] by D2,A3;
      end;
    suppose j = 0 & j = len F - 1; then
      0 + 1 = p + 1 by BINOM:def 7;
      hence P[j+1];
      end;
    suppose j = 0 & j > len F - 1;
      hence P[j+1] by C1;
      end;
    suppose D1: 0 < j & j < len F - 1; then
      D3: j + 1 < len F - 1 + 1 by XREAL_1:6;
      j + 1 > 0 + 1 by D1,XREAL_1:8;
      then F.(j+1) = 0.R by D3,A1;
      then fp.(j+1) = a|^p + 0.R by A4,C2,C1,A0;
      hence P[j+1] by D3;
      end;
    suppose 0 < j & j = len F - 1;
      hence P[j+1] by A0,A4,C2;
      end;
    suppose 0 < j & j > len F - 1;
      then j >= len F - 1 + 1 by INT_1:7;
      hence P[j+1] by C1;
      end;
    end;
I: for j being Element of NAT st 0 <= j & j <= len F holds P[j]
   from INT_1:sch 7(IA,IS);
H: len F = p + 1 by BINOM:def 7;
p is Element of NAT by ORDINAL1:def 12;
hence (a+b)|^p = Sum((a,b) In_Power p) by BINOM:25
              .= a|^p + b|^p by I,A2,H;
end;
