
theorem T4a:
for R being Ring
for a,b being Element of R holds a|R + b|R = (a + b)|R
proof
let R be Ring, a,b be Element of R;
set p = (a|R) + (b|R), q = (a + b)|R;
A: dom p = NAT by FUNCT_2:def 1 .= dom q by FUNCT_2:def 1;
now let x be object;
  assume x in dom q;
  then reconsider i = x as Element of NAT;
  B: p.i = (a|R).i + (b|R).i by NORMSP_1:def 2;
  per cases;
  suppose S: i = 0;
    hence q.x = a + b by Th28
             .= (a|R).i + b by S,Th28
             .= p.x by B,S,Th28;
    end;
  suppose S: i <> 0;
    hence q.x = 0.R by Th28
             .= (a|R).i + 0.R by S,Th28
             .= p.x by B,S,Th28;
    end;
  end;
hence thesis by A,FUNCT_1:2;
end;
