
theorem pr5:
for S being Ring,
    R being Subring of S
for x being Element of S, x1 being Element of R
for n being Element of NAT st x = x1 holds x|^n = x1|^n
proof
let S be Ring, R be Subring of S;
let x be Element of S, x1 be Element of R; let n be Element of NAT;
assume AS: x = x1;
defpred P[Nat] means
  for x be Element of S, x1 be Element of R st x=x1 holds x|^($1) = x1|^($1);
now let x be Element of S, x1 be Element of R;
  assume x = x1;
  thus x|^0 = 1_S by BINOM:8 .= 1_R by C0SP1:def 3 .= x1|^0 by BINOM:8;
  end; then
IA: P[0];
A: the multF of R = (the multF of S)||the carrier of R by C0SP1:def 3;
IS: now let k be Nat;
    assume IV: P[k];
    now let x be Element of S, x1 be Element of R;
      assume AS: x = x1; then
      C: x|^k = x1|^k by IV;
      B: [x1|^k,x1] in
               [:the carrier of R,the carrier of R:] by ZFMISC_1:def 2;
      thus x|^(k+1) = x|^k *x|^1 by BINOM:10
              .= x|^k * x by BINOM:8
              .= x1|^k * x1 by AS,A,B,C,FUNCT_1:49
              .= x1|^k * x1|^1 by BINOM:8
              .= x1|^(k+1) by BINOM:10;
      end;
    hence P[k+1];
    end;
for k being Nat holds P[k] from NAT_1:sch 2(IA,IS);
hence thesis by AS;
end;
