
theorem
for p being Prime
for R being p-characteristic commutative Ring
for a,b being Element of R
for i being Nat holds (a + b)|^(p|^i) = a|^(p|^i) + b|^(p|^i)
proof
let p be Prime, R be p-characteristic commutative Ring;
let a,b be Element of R, i be Nat;
defpred P[Nat] means (a + b)|^(p|^($1)) = a|^(p|^($1)) + b|^(p|^($1));
(a + b)|^(p|^0)
     = (a + b)|^1 by NEWTON:4
    .= a + b by BINOM:8
    .= a|^1 + b by BINOM:8
    .= a|^(p|^0) + b by NEWTON:4
    .= a|^(p|^0) + b|^1 by BINOM:8
    .= a|^(p|^0) + b|^(p|^0) by NEWTON:4; then
IA: P[0];
IS: now let k be Nat;
    assume IV: P[k];
    (a + b)|^(p|^(k+1))
        = (a + b)|^((p|^k)*p) by NEWTON:6
       .= ((a + b)|^(p|^k))|^p by BINOM:11
       .= (a|^(p|^k))|^p + (b|^(p|^k))|^p by IV,fresh
       .= a|^((p|^k)*p) + (b|^(p|^k))|^p by BINOM:11
       .= a|^((p|^k)*p) + b|^((p|^k)*p) by BINOM:11
       .= a|^(p|^(k+1)) + b|^((p|^k)*p) by NEWTON:6
       .= a|^(p|^(k+1)) + b|^(p|^(k+1)) by NEWTON:6;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis;
end;
