
theorem SQsub:
for R being Ring
for S being Subring of R holds SQ S c= SQ R
proof
let R be Ring, S be Subring of R;
AS4: the carrier of S c= the carrier of R by C0SP1:def 3;
let o be object;
  assume o in SQ S;
  then consider a being Element of S such that A: o = a & a is square;
  consider b being Element of S such that B: a = b^2 by A;
  reconsider a1 = a, b1 = b as Element of R by AS4;
  dom(the multF of S) = [:the carrier of S,the carrier of S:]
     by FUNCT_2:def 1;
  then B6: [b,b] in dom(the multF of S);
  a1 = ((the multF of R)||(the carrier of S)).(b,b) by C0SP1:def 3,B
    .= b1^2 by B6,FUNCT_1:49;
  then a1 is square;
  hence o in SQ R by A;
end;
