
theorem Xm:
for n being non trivial Nat
for R being Ring
for S being RingExtension of R holds X^(n,R) = X^(n,S)
proof
let n be non trivial Nat, F be Ring, E be RingExtension of F;
set q = X^(n,F), r = X^(n,E);
H1: F is Subring of E by FIELD_4:def 1; then
H2: 1.E = 1.F by C0SP1:def 3;
now let o be object;
   assume o in NAT;
   then reconsider m = o as Nat;
   per cases;
   suppose A: m = 1; then
     q.m = -1.F by Lm10 .= -1.E by H1,H2,FIELD_6:17 .= r.m by A,Lm10;
     hence q.o = r.o;
     end;
   suppose A: m = n; then
     q.m = 1.F by Lm10 .= 1.E by H1,C0SP1:def 3 .= r.m by A,Lm10;
     hence q.o = r.o;
     end;
   suppose A: m <> 1 & m <> n; then
     q.m = 0.F by Lm11 .= 0.E by H1,C0SP1:def 3 .= r.m by A,Lm11;
     hence q.o = r.o;
     end;
   end;
hence thesis;
end;
